Partial Diff Eq & Complex Ans

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### MODULE 1: PARTIAL DIFFERENTIAL EQUATIONS 1. **Form a partial differential equation from the relation $z=(x+y)f(x^2 - y^2)$.** * Let $u = x^2 - y^2$. Then $z = (x+y)f(u)$. * $p = \frac{\partial z}{\partial x} = f(u) + (x+y)f'(u) \cdot 2x$ * $q = \frac{\partial z}{\partial y} = f(u) - (x+y)f'(u) \cdot 2y$ * $yp - xq = y f(u) + 2xy(x+y)f'(u) - x f(u) + 2xy(x+y)f'(u)$ * $yp - xq = (y-x)f(u) + 4xy(x+y)f'(u)$ * This is not directly eliminating $f$. A common approach for this type of problem is to use: * $\frac{p}{x+y} = \frac{f(u)}{x+y} + 2xf'(u)$ * $\frac{q}{x+y} = \frac{f(u)}{x+y} - 2yf'(u)$ * $(\frac{p}{x+y} - \frac{f(u)}{x+y}) / (2x) = f'(u)$ * $(\frac{q}{x+y} - \frac{f(u)}{x+y}) / (-2y) = f'(u)$ * Equating: $\frac{p - f(u)}{2x(x+y)} = \frac{q - f(u)}{-2y(x+y)}$ * $-y(p-f(u)) = x(q-f(u))$ * $-yp + yf(u) = xq - xf(u)$ * $-yp + xq = -(x+y)f(u)$ * Substitute $f(u) = \frac{z}{x+y}$: * $-yp + xq = -(x+y)\frac{z}{x+y}$ * $\boxed{xq - yp = -z}$ or $\boxed{yp - xq = z}$ 2. **Solve $\frac{\partial^3 z}{\partial x^2 \partial y} = \cos(3x+4y)$.** * Integrate with respect to $y$: $\frac{\partial^2 z}{\partial x^2} = \int \cos(3x+4y) dy = \frac{1}{4}\sin(3x+4y) + F(x)$ * Integrate with respect to $x$: $\frac{\partial z}{\partial x} = \int (\frac{1}{4}\sin(3x+4y) + F(x)) dx = -\frac{1}{12}\cos(3x+4y) + \int F(x) dx + G(y)$ Let $\int F(x) dx = \phi(x)$ $\frac{\partial z}{\partial x} = -\frac{1}{12}\cos(3x+4y) + \phi(x) + G(y)$ * Integrate with respect to $x$ again: $z = \int (-\frac{1}{12}\cos(3x+4y) + \phi(x) + G(y)) dx$ $\boxed{z = -\frac{1}{36}\sin(3x+4y) + \Phi(x) + xG(y) + H(y)}$ (where $\Phi(x) = \int \phi(x) dx$) 3. **Derive a partial differential equation from the relation $z=(x+y)f(x^2 - y^2)$.** * (This is the same as question 1, result is $\boxed{xq - yp = -z}$ or $\boxed{yp - xq = z}$) 4. **Solve using direct integration: $\frac{\partial^2 u}{\partial x \partial t} = e^{-t} \cos x$.** * Integrate with respect to $x$: $\frac{\partial u}{\partial t} = \int e^{-t} \cos x dx = e^{-t} \sin x + F(t)$ * Integrate with respect to $t$: $u = \int (e^{-t} \sin x + F(t)) dt$ $\boxed{u = -e^{-t} \sin x + \Phi(t) + G(x)}$ (where $\Phi(t) = \int F(t) dt$) 5. **Form the PDE for the equation $z=f(x^2 - y^2)$ where f is an arbitrary function.** * Let $u = x^2 - y^2$. Then $z = f(u)$. * $p = \frac{\partial z}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x} = f'(u) \cdot 2x$ * $q = \frac{\partial z}{\partial y} = f'(u) \cdot \frac{\partial u}{\partial y} = f'(u) \cdot (-2y)$ * From $p = 2xf'(u)$, $f'(u) = \frac{p}{2x}$ * From $q = -2yf'(u)$, $f'(u) = -\frac{q}{2y}$ * Equating $f'(u)$: $\frac{p}{2x} = -\frac{q}{2y}$ * $py = -qx$ * $\boxed{py + qx = 0}$ 6. **Solve $\frac{\partial^2 z}{\partial x^2} + z = 0$ given that when $x=0$, $z=e^y$ and $\frac{\partial z}{\partial x}=1$.** * This is a second-order linear ordinary differential equation in $x$, treating $y$ as a constant. * The characteristic equation is $m^2 + 1 = 0 \implies m = \pm i$. * The general solution is $z(x,y) = A(y)\cos x + B(y)\sin x$. * Apply boundary conditions: * When $x=0$, $z=e^y$: $e^y = A(y)\cos(0) + B(y)\sin(0) \implies A(y) = e^y$. * So, $z(x,y) = e^y \cos x + B(y)\sin x$. * Now find $\frac{\partial z}{\partial x}$: $\frac{\partial z}{\partial x} = -e^y \sin x + B(y)\cos x$. * When $x=0$, $\frac{\partial z}{\partial x}=1$: $1 = -e^y \sin(0) + B(y)\cos(0) \implies B(y) = 1$. * Therefore, the solution is $\boxed{z = e^y \cos x + \sin x}$. 7. **Find the partial differential equation by eliminating arbitrary functions f and g from $z=f(x)+g(y)$.** * $p = \frac{\partial z}{\partial x} = f'(x)$ * $q = \frac{\partial z}{\partial y} = g'(y)$ * Differentiate $p$ with respect to $y$: $\frac{\partial p}{\partial y} = \frac{\partial^2 z}{\partial y \partial x} = 0$, since $f'(x)$ is a function of $x$ only. * Differentiate $q$ with respect to $x$: $\frac{\partial q}{\partial x} = \frac{\partial^2 z}{\partial x \partial y} = 0$, since $g'(y)$ is a function of $y$ only. * $\boxed{\frac{\partial^2 z}{\partial x \partial y} = 0}$ or $\boxed{r = 0}$ or $\boxed{s = 0}$ 8. **Solve $\frac{\partial^2 z}{\partial x^2} = xy$.** * Integrate with respect to $x$: $\frac{\partial z}{\partial x} = \int (xy) dx = \frac{x^2 y}{2} + F(y)$ * Integrate with respect to $x$ again: $z = \int (\frac{x^2 y}{2} + F(y)) dx = \frac{x^3 y}{6} + xF(y) + G(y)$ * $\boxed{z = \frac{x^3 y}{6} + xF(y) + G(y)}$ (where $F(y)$ and $G(y)$ are arbitrary functions of $y$) 9. **Form the partial differential equation by eliminating arbitrary function from the relation $z=x-y+f(x^2 + y^2)$.** * Let $u = x^2 + y^2$. Then $z = x-y+f(u)$. * $p = \frac{\partial z}{\partial x} = 1 + f'(u) \cdot 2x$ * $q = \frac{\partial z}{\partial y} = -1 + f'(u) \cdot 2y$ * From $p-1 = 2xf'(u)$, $f'(u) = \frac{p-1}{2x}$ * From $q+1 = 2yf'(u)$, $f'(u) = \frac{q+1}{2y}$ * Equating $f'(u)$: $\frac{p-1}{2x} = \frac{q+1}{2y}$ * $y(p-1) = x(q+1)$ * $yp - y = xq + x$ * $\boxed{yp - xq = x+y}$ 10. **Solve by direct integration: $\frac{\partial^2 z}{\partial x \partial y} = \sin(3x+4y)$.** * Integrate with respect to $y$: $\frac{\partial z}{\partial x} = \int \sin(3x+4y) dy = -\frac{1}{4}\cos(3x+4y) + F(x)$ * Integrate with respect to $x$: $z = \int (-\frac{1}{4}\cos(3x+4y) + F(x)) dx$ $\boxed{z = -\frac{1}{12}\sin(3x+4y) + \int F(x) dx + G(y)}$ Or, $\boxed{z = -\frac{1}{12}\sin(3x+4y) + \phi(x) + G(y)}$ (where $\phi(x)$ and $G(y)$ are arbitrary functions) 11. **Derive partial differential equation from $z=f(x+2t)+g(x-2t)$.** * Let $u = x+2t$ and $v = x-2t$. Then $z = f(u) + g(v)$. * $\frac{\partial z}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x} + g'(v) \cdot \frac{\partial v}{\partial x} = f'(u) + g'(v)$ * $\frac{\partial^2 z}{\partial x^2} = f''(u) + g''(v)$ * $\frac{\partial z}{\partial t} = f'(u) \cdot \frac{\partial u}{\partial t} + g'(v) \cdot \frac{\partial v}{\partial t} = 2f'(u) - 2g'(v)$ * $\frac{\partial^2 z}{\partial t^2} = 2f''(u) \cdot 2 - 2g''(v) \cdot (-2) = 4f''(u) + 4g''(v) = 4(f''(u) + g''(v))$ * Substitute $\frac{\partial^2 z}{\partial x^2}$: * $\boxed{\frac{\partial^2 z}{\partial t^2} = 4 \frac{\partial^2 z}{\partial x^2}}$ (This is a 1D wave equation) 12. **Solve $xp+yq=z$.** (This is a Lagrange's method problem for a linear PDE) * The auxiliary equations are $\frac{dx}{x} = \frac{dy}{y} = \frac{dz}{z}$. * From $\frac{dx}{x} = \frac{dy}{y}$: $\ln|x| = \ln|y| + \ln|c_1| \implies \frac{x}{y} = c_1$. * From $\frac{dy}{y} = \frac{dz}{z}$: $\ln|y| = \ln|z| + \ln|c_2| \implies \frac{y}{z} = c_2$. * The general solution is $\Phi(c_1, c_2) = 0$ or $c_2 = F(c_1)$. * $\boxed{\frac{y}{z} = F(\frac{x}{y})}$ or $\boxed{z = y G(\frac{x}{y})}$ 13. **Find the differential equation of all spheres of fixed radius having their centres in the xy-plane.** * Let the fixed radius be $r$. * The center is $(a, b, 0)$. * The equation of such a sphere is $(x-a)^2 + (y-b)^2 + z^2 = r^2$. * Differentiating with respect to $x$: $2(x-a) + 2z \frac{\partial z}{\partial x} = 0 \implies x-a = -zp$ (where $p = \frac{\partial z}{\partial x}$) * Differentiating with respect to $y$: $2(y-b) + 2z \frac{\partial z}{\partial y} = 0 \implies y-b = -zq$ (where $q = \frac{\partial z}{\partial y}$) * Substitute $x-a$ and $y-b$ back into the sphere equation: $(-zp)^2 + (-zq)^2 + z^2 = r^2$ $z^2 p^2 + z^2 q^2 + z^2 = r^2$ $\boxed{z^2(p^2 + q^2 + 1) = r^2}$ 14. **Solve $\frac{\partial^2 z}{\partial x \partial y} = \frac{x}{y} + a$.** * Integrate with respect to $y$: $\frac{\partial z}{\partial x} = \int (\frac{x}{y} + a) dy = x\ln|y| + ay + F(x)$ * Integrate with respect to $x$: $z = \int (x\ln|y| + ay + F(x)) dx$ $\boxed{z = \frac{x^2}{2}\ln|y| + axy + \phi(x) + G(y)}$ (where $\phi(x)$ and $G(y)$ are arbitrary functions) 15. **Form the partial differential equation by eliminating arbitrary constant from the relation $2z=\frac{x^2}{a^2} + \frac{y^2}{b^2}$.** * $p = \frac{\partial z}{\partial x} = \frac{2x}{2a^2} = \frac{x}{a^2}$ * $q = \frac{\partial z}{\partial y} = \frac{2y}{2b^2} = \frac{y}{b^2}$ * From $p=\frac{x}{a^2} \implies a^2 = \frac{x}{p}$ * From $q=\frac{y}{b^2} \implies b^2 = \frac{y}{q}$ * Substitute $a^2$ and $b^2$ into the original equation: $2z = \frac{x^2}{x/p} + \frac{y^2}{y/q}$ $2z = xp + yq$ $\boxed{xp + yq = 2z}$ 16. **Solve $py+xq-xz=0$.** (This is a Lagrange's method problem) * The auxiliary equations are $\frac{dx}{y} = \frac{dy}{x} = \frac{dz}{xz}$. * From $\frac{dx}{y} = \frac{dy}{x}$: $x dx = y dy \implies \int x dx = \int y dy \implies \frac{x^2}{2} = \frac{y^2}{2} + c \implies x^2 - y^2 = c_1$. * Now use $dx, dy, dz$ and the multipliers $x, -y, 0$ on $\frac{dx}{y} = \frac{dy}{x}$: $\frac{x dx - y dy}{xy - yx} = \frac{dz}{xz}$. This doesn't work well due to the denominator. * Let's use $\frac{dy}{x} = \frac{dz}{xz}$. $\frac{dy}{1} = \frac{dz}{z} \implies \int dy = \int \frac{dz}{z} \implies y = \ln|z| + c' \implies e^{y-c'} = z \implies z = c_2 e^y$. This is if $x \ne 0$. * If we use $\frac{dy}{x} = \frac{dz}{xz}$, we can simplify by $x$: $\frac{dy}{1} = \frac{dz}{z}$. $\ln|y| = \ln|z| + \ln|c_2| \implies \frac{y}{z} = c_2$. * So the solutions are $c_1 = x^2 - y^2$ and $c_2 = y/z$. * The general solution is $\Phi(c_1, c_2) = 0$ or $c_2 = F(c_1)$. * $\boxed{\frac{y}{z} = F(x^2 - y^2)}$ or $\boxed{z = \frac{y}{F(x^2-y^2)}}$ 17. **Find the differential equation of all spheres whose centers lie on z-axis.** * The center is $(0, 0, c)$ and let radius be $r$. * The equation of such a sphere is $x^2 + y^2 + (z-c)^2 = r^2$. * Differentiating with respect to $x$: $2x + 2(z-c) \frac{\partial z}{\partial x} = 0 \implies x + (z-c)p = 0 \implies z-c = -\frac{x}{p}$ * Differentiating with respect to $y$: $2y + 2(z-c) \frac{\partial z}{\partial y} = 0 \implies y + (z-c)q = 0 \implies z-c = -\frac{y}{q}$ * Equating the expressions for $z-c$: $-\frac{x}{p} = -\frac{y}{q}$ $xq = yp$ $\boxed{xq - yp = 0}$ 18. **Solve $\frac{\partial^2 z}{\partial y \partial x} = \frac{x}{y} + a$.** (This is identical to question 14) * The solution is $\boxed{z = \frac{x^2}{2}\ln|y| + axy + \phi(x) + G(y)}$. ### MODULE 2: APPLICATIONS OF PDEs 1. **Write any three assumptions involved in the derivation of one dimensional wave equation.** * The string is perfectly flexible (no resistance to bending). * The tension in the string is constant and great compared to the weight of the string. * The string is homogeneous (uniform mass density). * The displacement of the string is small, so that the slope $\frac{\partial y}{\partial x}$ is small and its square can be neglected. * Motion is in a plane transverse to the x-axis (no longitudinal displacement). 2. **Find the steady state temperature distribution in a rod of length 25 cm, if the ends of the rod are kept at $20^\circ C$ and $70^\circ C$.** * For steady-state, the 1D heat equation is $\frac{\partial^2 u}{\partial x^2} = 0$. * Integrating twice gives $u(x) = Ax + B$. * Boundary conditions: * $u(0) = 20^\circ C \implies A(0) + B = 20 \implies B=20$. * $u(25) = 70^\circ C \implies A(25) + B = 70$. * $25A + 20 = 70 \implies 25A = 50 \implies A=2$. * The steady-state temperature distribution is $\boxed{u(x) = 2x + 20}$. 3. **Solve $2z=xp+yq$.** (This is a Clairaut's form or a Lagrange's method problem) * This PDE is of the form $z = Px + Qy + R$. * The auxiliary equations are $\frac{dx}{x} = \frac{dy}{y} = \frac{dz}{2z}$. * From $\frac{dx}{x} = \frac{dy}{y}$: $\ln|x| = \ln|y| + \ln|c_1| \implies \frac{x}{y} = c_1$. * From $\frac{dy}{y} = \frac{dz}{2z}$: $\ln|y| = \frac{1}{2}\ln|z| + \ln|c_2| \implies y^2 = c_2 z$. * The general solution is $\Phi(\frac{x}{y}, \frac{y^2}{z}) = 0$ or $\frac{y^2}{z} = F(\frac{x}{y})$. * $\boxed{z = \frac{y^2}{F(x/y)}}$ or $\boxed{z = y^2 G(x/y)}$. 4. **Write any three assumptions in deriving one dimensional heat equation.** * The material of the rod is homogeneous and its thermal properties (density, specific heat, thermal conductivity) are constant. * The rod is perfectly insulated laterally, so heat flow is only along the axis of the rod (1D flow). * The cross-section of the rod is uniform. * There are no heat sources or sinks within the rod (unless a source term is included in the equation). * The temperature is uniform across any given cross-section. 5. **Write the conditions in which a tightly stretched string of length l with fixed end is initially in equilibrium position and is set vibrating by giving each point a velocity $v_0 \sin \frac{3\pi x}{l}$.** * **Boundary Conditions:** * $u(0, t) = 0$ for all $t > 0$ (Fixed at $x=0$) * $u(l, t) = 0$ for all $t > 0$ (Fixed at $x=l$) * **Initial Conditions:** * $u(x, 0) = 0$ for $0 \le x \le l$ (Initially in equilibrium position, i.e., zero initial displacement) * $\frac{\partial u}{\partial t}(x, 0) = v_0 \sin \frac{3\pi x}{l}$ for $0 \le x \le l$ (Initial velocity distribution) 6. **Write down the possible solutions of one dimensional heat equation.** * The 1D heat equation is $\frac{\partial u}{\partial t} = c^2 \frac{\partial^2 u}{\partial x^2}$. * Using separation of variables $u(x,t) = X(x)T(t)$, it leads to three possible forms: * Case 1: If $\lambda = 0$, then $X(x) = A_1 x + B_1$ and $T(t) = C_1$. So $u(x,t) = (A_1 x + B_1)C_1 = Ax + B$. * Case 2: If $\lambda = \mu^2 > 0$, then $X(x) = A_2 e^{\mu x} + B_2 e^{-\mu x}$ and $T(t) = C_2 e^{c^2 \mu^2 t}$. So $u(x,t) = (A_2 e^{\mu x} + B_2 e^{-\mu x}) e^{c^2 \mu^2 t}$. * Case 3: If $\lambda = -\mu^2 0$, then $X(x) = A_2 e^{\mu x} + B_2 e^{-\mu x}$ and $T(t) = C_2 e^{c\mu t} + D_2 e^{-c\mu t}$. So $u(x,t) = (A_2 e^{\mu x} + B_2 e^{-\mu x})(C_2 e^{c\mu t} + D_2 e^{-c\mu t})$. * Case 3: If $\lambda = -\mu^2 ### MODULE 3: COMPLEX FUNCTIONS 1. **Determine whether $w=\cos z$ is analytic.** * Let $z = x+iy$. Then $\cos z = \cos(x+iy) = \cos x \cos(iy) - \sin x \sin(iy) = \cos x \cosh y - i \sin x \sinh y$. * So, $u(x,y) = \cos x \cosh y$ and $v(x,y) = -\sin x \sinh y$. * Find the partial derivatives: * $u_x = -\sin x \cosh y$ * $u_y = \cos x \sinh y$ * $v_x = -\cos x \sinh y$ * $v_y = -\sin x \cosh y$ * Check Cauchy-Riemann equations: * $u_x = v_y \implies -\sin x \cosh y = -\sin x \cosh y$ (True) * $u_y = -v_x \implies \cos x \sinh y = -(-\cos x \sinh y) \implies \cos x \sinh y = \cos x \sinh y$ (True) * Since the Cauchy-Riemann equations are satisfied and the partial derivatives are continuous everywhere, $w=\cos z$ is $\boxed{\text{analytic everywhere (entire)}}$. 2. **Check whether the function $xy^2$ is the real part of an analytic function.** * Let $u(x,y) = xy^2$. For $u$ to be the real part of an analytic function, it must be harmonic. * Find second partial derivatives: * $u_x = y^2$ * $u_{xx} = 0$ * $u_y = 2xy$ * $u_{yy} = 2x$ * Check Laplace's equation: $u_{xx} + u_{yy} = 0$. * $0 + 2x = 0 \implies 2x = 0$. * This is only true if $x=0$. Since it's not true for all $x$, $u(x,y) = xy^2$ is not harmonic for all $(x,y)$. * Therefore, $xy^2$ is $\boxed{\text{not the real part of an analytic function}}$. 3. **Show that an analytic function $f(z)=u+iv$ is constant if its real part is constant.** * Given $f(z) = u+iv$ is analytic and $u(x,y) = C$ (constant). * Since $u$ is constant, its partial derivatives are zero: $u_x = 0$ and $u_y = 0$. * For $f(z)$ to be analytic, it must satisfy the Cauchy-Riemann equations: * $u_x = v_y$ * $u_y = -v_x$ * From $u_x = 0$, we have $v_y = 0$. * From $u_y = 0$, we have $0 = -v_x$, which means $v_x = 0$. * Since $v_x = 0$ and $v_y = 0$, $v(x,y)$ must also be a constant. * As both $u$ and $v$ are constants, $f(z) = u+iv$ is a $\boxed{\text{constant function}}$. 4. **Show that the function $u=\sin x \cosh y$ is harmonic.** * A function is harmonic if it satisfies Laplace's equation: $u_{xx} + u_{yy} = 0$. * Given $u(x,y) = \sin x \cosh y$. * Find partial derivatives: * $u_x = \cos x \cosh y$ * $u_{xx} = -\sin x \cosh y$ * $u_y = \sin x \sinh y$ * $u_{yy} = \sin x \cosh y$ * Substitute into Laplace's equation: $u_{xx} + u_{yy} = (-\sin x \cosh y) + (\sin x \cosh y) = 0$. * Since $u_{xx} + u_{yy} = 0$, the function $u=\sin x \cosh y$ is $\boxed{\text{harmonic}}$. 5. **Find the real part and imaginary part of the function $f(z)=5z^2 - 12z + 3 + 2i$ and find their values at $z=4-3i$.** * Let $z = x+iy$. * $f(z) = 5(x+iy)^2 - 12(x+iy) + 3 + 2i$ * $f(z) = 5(x^2 - y^2 + 2ixy) - 12x - 12iy + 3 + 2i$ * $f(z) = 5x^2 - 5y^2 + 10ixy - 12x - 12iy + 3 + 2i$ * Group real and imaginary parts: * Real Part $u(x,y) = 5x^2 - 5y^2 - 12x + 3$ * Imaginary Part $v(x,y) = 10xy - 12y + 2$ * Now evaluate at $z=4-3i$, so $x=4, y=-3$: * $u(4,-3) = 5(4^2) - 5(-3)^2 - 12(4) + 3$ $u(4,-3) = 5(16) - 5(9) - 48 + 3$ $u(4,-3) = 80 - 45 - 48 + 3 = 35 - 48 + 3 = -13 + 3 = -10$ * $v(4,-3) = 10(4)(-3) - 12(-3) + 2$ $v(4,-3) = -120 + 36 + 2 = -84 + 2 = -82$ * Real part at $z=4-3i$ is $\boxed{-10}$. * Imaginary part at $z=4-3i$ is $\boxed{-82}$. 6. **Find the fixed points of the mapping $w=(a+ib)z^2$.** * A fixed point is a point where $w=z$. * So we need to solve $z = (a+ib)z^2$. * $z - (a+ib)z^2 = 0$ * $z(1 - (a+ib)z) = 0$ * This gives two possibilities: * Case 1: $z = 0$. This is always a fixed point. * Case 2: $1 - (a+ib)z = 0 \implies (a+ib)z = 1 \implies z = \frac{1}{a+ib}$. Rationalize the denominator: $z = \frac{1}{a+ib} \cdot \frac{a-ib}{a-ib} = \frac{a-ib}{a^2+b^2}$. * The fixed points are $\boxed{z=0}$ and $\boxed{z=\frac{a-ib}{a^2+b^2}}$ (provided $a^2+b^2 \ne 0$). 7. **Test the continuity at $z=0$ of $f(z)=\begin{cases} \frac{\operatorname{Im}(z)}{|z|} & z \ne 0 \\ 0 & z=0 \end{cases}$.** * For continuity at $z=0$, we need $\lim_{z \to 0} f(z) = f(0)$. * $f(0) = 0$. * Let $z \to 0$ along the path $y=mx$ (where $z=x+iy$). * $\operatorname{Im}(z) = y = mx$ * $|z| = \sqrt{x^2+y^2} = \sqrt{x^2+(mx)^2} = \sqrt{x^2(1+m^2)} = |x|\sqrt{1+m^2}$ * As $z \to 0$, $x \to 0$. * $\lim_{x \to 0} \frac{mx}{|x|\sqrt{1+m^2}}$. * If $x > 0$, $\lim_{x \to 0^+} \frac{mx}{x\sqrt{1+m^2}} = \frac{m}{\sqrt{1+m^2}}$. * If $x 0$): $\lim_{x \to 0^+} \frac{x}{x} = 1$. * Along the positive y-axis ($x=0, y>0$): $\lim_{y \to 0^+} \frac{iy}{y} = i$. * Since the limit depends on the path, $\lim_{z \to 0} f(z)$ does not exist. * Therefore, the function is $\boxed{\text{not continuous at } z=0}$. 17. **Determine 'a' so that $u=e^{-\pi x} \cos ay$ is harmonic.** * For $u$ to be harmonic, it must satisfy Laplace's equation $u_{xx} + u_{yy} = 0$. * Given $u(x,y) = e^{-\pi x} \cos ay$. * Find partial derivatives: * $u_x = -\pi e^{-\pi x} \cos ay$ * $u_{xx} = (-\pi)(-\pi) e^{-\pi x} \cos ay = \pi^2 e^{-\pi x} \cos ay$ * $u_y = -a e^{-\pi x} \sin ay$ * $u_{yy} = -a(a) e^{-\pi x} \cos ay = -a^2 e^{-\pi x} \cos ay$ * Substitute into Laplace's equation: $\pi^2 e^{-\pi x} \cos ay - a^2 e^{-\pi x} \cos ay = 0$ $e^{-\pi x} \cos ay (\pi^2 - a^2) = 0$ * For this to be true for all $x,y$ (and assuming $\cos ay \ne 0$), we must have: $\pi^2 - a^2 = 0 \implies a^2 = \pi^2 \implies \boxed{a = \pm \pi}$. 18. **Prove that an analytic function of constant real part is constant.** (Same as question 3) * See answer to question 3. ### MODULE 4: COMPLEX INTEGRATION & SERIES 1. **Using Cauchy's integral formula, Evaluate $\int_C \frac{z^2+1}{z^2-1}dz$ where C is the circle of unit radius with centre at $z=1$.** * The integrand is $f(z) = \frac{z^2+1}{(z-1)(z+1)}$. * The circle $C$ is $|z-1|=1$. This circle encloses the singularity $z=1$ but not $z=-1$. * We rewrite the integrand to fit Cauchy's formula $\int_C \frac{g(z)}{z-z_0} dz = 2\pi i g(z_0)$: $\int_C \frac{(z^2+1)/(z+1)}{z-1} dz$. * Here, $g(z) = \frac{z^2+1}{z+1}$ and $z_0 = 1$. * $g(z)$ is analytic inside and on $C$. * Evaluate $g(z_0)$: $g(1) = \frac{1^2+1}{1+1} = \frac{2}{2} = 1$. * Using Cauchy's Integral Formula: $\int_C \frac{z^2+1}{z^2-1} dz = 2\pi i \cdot g(1) = 2\pi i \cdot 1 = \boxed{2\pi i}$. 2. **Find the Taylor's series of $\frac{1}{z}$ about $z=1$.** * The Taylor series of a function $f(z)$ about $z=z_0$ is given by $\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$. * Here, $f(z) = \frac{1}{z}$ and $z_0=1$. * We can use a geometric series expansion: $\frac{1}{z} = \frac{1}{1+(z-1)} = \frac{1}{1 - (-(z-1))}$. * Using the formula $\frac{1}{1-r} = \sum_{n=0}^\infty r^n$ for $|r| 4$). * Since the integrand $f(z) = \frac{e^z}{z-5}$ is analytic at all points inside and on $C$, by Cauchy's Integral Theorem, the integral is 0. * $\boxed{0}$. 14. **Find the Taylor series expansion of $e^z$ about $z=\pi$.** * The Taylor series of $f(z)$ about $z=z_0$ is $\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$. * Here, $f(z) = e^z$ and $z_0=\pi$. * The derivatives of $e^z$ are all $e^z$. * So, $f^{(n)}(z) = e^z$. * At $z_0=\pi$, $f^{(n)}(\pi) = e^\pi$. * The Taylor series is $\sum_{n=0}^\infty \frac{e^\pi}{n!}(z-\pi)^n$. * This can also be written as $e^\pi \sum_{n=0}^\infty \frac{(z-\pi)^n}{n!}$. * This series is valid for all $z$. * $\boxed{e^z = e^\pi \left(1 + (z-\pi) + \frac{(z-\pi)^2}{2!} + \frac{(z-\pi)^3}{3!} + \dots \right)}$. 15. **Find the residue of $\frac{z^2}{z^2+a^2}$ at $z=ia$.** * The denominator $z^2+a^2 = (z-ia)(z+ia)$. * The function has simple poles at $z=ia$ and $z=-ia$. * We need the residue at $z=ia$. * Let $f(z) = \frac{z^2}{(z-ia)(z+ia)}$. * This is a simple pole, so $\operatorname{Res}(f, ia) = \lim_{z \to ia} (z-ia)f(z)$. * $\operatorname{Res}(f, ia) = \lim_{z \to ia} (z-ia) \frac{z^2}{(z-ia)(z+ia)} = \lim_{z \to ia} \frac{z^2}{z+ia}$. * $\operatorname{Res}(f, ia) = \frac{(ia)^2}{ia+ia} = \frac{-a^2}{2ia} = \frac{ia^2}{2a} = \boxed{\frac{ia}{2}}$. 16. **Give examples of removable singularity, pole and essential singularity.** * **Removable Singularity:** * Example: $f(z) = \frac{\sin z}{z}$ at $z=0$. * $\lim_{z \to 0} \frac{\sin z}{z} = 1$. Since the limit exists and is finite, the singularity at $z=0$ can be removed by defining $f(0)=1$. * **Pole:** * Example: $f(z) = \frac{1}{z^3}$ at $z=0$. * The Laurent series is already in form, with the highest negative power of $z$ being $-3$. This is a pole of order 3. * Example: $f(z) = \tan z = \frac{\sin z}{\cos z}$ at $z=\frac{\pi}{2}$. $\cos(\frac{\pi}{2})=0$ but $\sin(\frac{\pi}{2})=1 \ne 0$. This is a simple pole. * **Essential Singularity:** * Example: $f(z) = e^{1/z}$ at $z=0$. * The Laurent series $e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \dots$ has infinitely many terms in its principal part. 17. **Find the value of the integral $\int_C z dz$ where C is the part of the unit circle from $-i$ to $i$.** * The function $f(z)=z$ is entire (analytic everywhere). * Thus, the integral is path-independent. We can use an antiderivative. * An antiderivative of $z$ is $F(z) = \frac{z^2}{2}$. * The integral is $F(i) - F(-i)$. * $\int_C z dz = \frac{i^2}{2} - \frac{(-i)^2}{2} = \frac{-1}{2} - \frac{-1}{2} = \boxed{0}$. * Alternatively, parameterize the path. The part of the unit circle from $-i$ to $i$ can be $z(t) = e^{it}$ for $t \in [-\pi/2, \pi/2]$. * $dz = i e^{it} dt$. * $\int_{-\pi/2}^{\pi/2} e^{it} (i e^{it}) dt = \int_{-\pi/2}^{\pi/2} i e^{2it} dt = i \left[ \frac{e^{2it}}{2i} \right]_{-\pi/2}^{\pi/2} = \frac{1}{2} [e^{2it}]_{-\pi/2}^{\pi/2}$ * $= \frac{1}{2} (e^{i\pi} - e^{-i\pi}) = \frac{1}{2} (-1 - (-1)) = \frac{1}{2}(-1+1) = \boxed{0}$. 18. **Evaluate $\oint_C \frac{\cos z}{z^2+9} dz$ where C is $|z|=1$.** * The denominator $z^2+9 = (z-3i)(z+3i)$. * The singularities are $z=3i$ and $z=-3i$. * The contour $C$ is the circle $|z|=1$. * Both singularities $z=3i$ (since $|3i|=3 > 1$) and $z=-3i$ (since $|-3i|=3 > 1$) lie outside the contour $C$. * Since the integrand $f(z) = \frac{\cos z}{z^2+9}$ is analytic inside and on $C$, by Cauchy's Integral Theorem, the integral is 0. * $\boxed{0}$. 19. **Find the Laurent series expansion of $\frac{1}{1-z^2}$ about $z=0$.** * We can use the geometric series formula $\frac{1}{1-r} = \sum_{n=0}^\infty r^n$. * Here, if we let $r=z^2$, we get: $\frac{1}{1-z^2} = \sum_{n=0}^\infty (z^2)^n = \sum_{n=0}^\infty z^{2n}$. * This expansion is a Maclaurin series, which is a special type of Laurent series (where the principal part is zero). * This series is valid for $|z^2| \frac{1}{2}$). * Rewrite the integrand to use Cauchy's Integral Formula around $z=0$: $\frac{e^z}{z(z-1)} = \frac{e^z/(z-1)}{z}$. * Here $g(z) = \frac{e^z}{z-1}$. This function is analytic inside and on $C$. * $z_0=0$. * Evaluate $g(z_0)$: $g(0) = \frac{e^0}{0-1} = \frac{1}{-1} = -1$. * By Cauchy's Integral Formula: $\oint_C \frac{e^z}{z(z-1)} dz = 2\pi i \cdot g(0) = 2\pi i \cdot (-1) = \boxed{-2\pi i}$. 23. **Find the Taylor series expansion of $\frac{1}{z}$ about $z=2$.** * The Taylor series of $f(z)$ about $z=z_0$ is $\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$. * Here, $f(z) = \frac{1}{z}$ and $z_0=2$. * We can use a geometric series approach: $\frac{1}{z} = \frac{1}{2 + (z-2)} = \frac{1}{2(1 + \frac{z-2}{2})} = \frac{1}{2} \frac{1}{1 - (-\frac{z-2}{2})}$. * Using $\frac{1}{1-r} = \sum_{n=0}^\infty r^n$: $\frac{1}{z} = \frac{1}{2} \sum_{n=0}^\infty \left(-\frac{z-2}{2}\right)^n = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n (z-2)^n}{2^n}$. $\boxed{\frac{1}{z} = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}} (z-2)^n}$. * This series is valid for $|-\frac{z-2}{2}| 2$. So $z=-2$ is outside $C$. * We only need to consider the residue at $z=2$. * Rewrite the integrand for Cauchy's Integral Formula around $z=2$: $\frac{\sin z}{(z-2)(z+2)} = \frac{\sin z/(z+2)}{z-2}$. * Here $g(z) = \frac{\sin z}{z+2}$. This function is analytic inside and on $C$. * $z_0=2$. * Evaluate $g(z_0)$: $g(2) = \frac{\sin 2}{2+2} = \frac{\sin 2}{4}$. * By Cauchy's Integral Formula: $\oint_C \frac{\sin z}{z^2-4} dz = 2\pi i \cdot g(2) = 2\pi i \cdot \frac{\sin 2}{4} = \boxed{\frac{\pi i \sin 2}{2}}$. 31. **Find the Laurent series expansion of $\frac{1}{1+z}$ about $z=0$.** * This is a Maclaurin series, which is a special case of a Laurent series. * Using the geometric series formula $\frac{1}{1-r} = \sum_{n=0}^\infty r^n$. * Here, let $r = -z$. * $\frac{1}{1+z} = \frac{1}{1-(-z)} = \sum_{n=0}^\infty (-z)^n = \sum_{n=0}^\infty (-1)^n z^n$. * This series is valid for $|-z| 2$, $z=2i$ is outside $C$. * For $z=-2i$: $|-2i-2| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} \approx 2.828$. Since $\sqrt{8} > 2$, $z=-2i$ is outside $C$. * Since both singularities lie outside the contour $C$, the integrand is analytic inside and on $C$. * By Cauchy's Integral Theorem, the integral is $\boxed{0}$. 35. **What type of singularity does $f(z)=e^{1/z}$ have?** * The singularity is at $z=0$. * The Laurent series expansion for $e^{1/z}$ about $z=0$ is: $e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \dots$ * This series has infinitely many terms in its principal part (terms with negative powers of $z$). * Therefore, $z=0$ is an $\boxed{\text{essential singularity}}$. 36. **Find the residue of $f(z)=\frac{\sin 2z}{z^2}$ at its pole.** * The singularity is at $z=0$. * We need to find the Laurent series expansion of $f(z)$ about $z=0$. * The Maclaurin series for $\sin w = w - \frac{w^3}{3!} + \frac{w^5}{5!} - \dots$. * Replace $w$ with $2z$: $\sin(2z) = (2z) - \frac{(2z)^3}{3!} + \frac{(2z)^5}{5!} - \dots = 2z - \frac{8z^3}{6} + \frac{32z^5}{120} - \dots$ * Now divide by $z^2$: $f(z) = \frac{1}{z^2} \left( 2z - \frac{8z^3}{6} + \dots \right)$ $f(z) = \frac{2}{z} - \frac{8z}{6} + \dots = \frac{2}{z} - \frac{4z}{3} + \dots$ * The highest negative power of $z$ is $z^{-1}$, so $z=0$ is a simple pole (pole of order 1). * The residue is the coefficient of $z^{-1}$, which is $2$. * The residue at $z=0$ is $\boxed{2}$. * (Note: If it were $\sin 2z / z^3$, the pole would be of order 2).