Real and Imaginary Parts of Complex Func

Cheatsheet Content

Complex Numbers Review A complex number $z$ is written as $z = x + iy$, where $x, y \in \mathbb{R}$. $x = \text{Re}(z)$ is the real part. $y = \text{Im}(z)$ is the imaginary part. $i$ is the imaginary unit, $i^2 = -1$. Polar form: $z = r(\cos \theta + i \sin \theta) = re^{i\theta}$. $r = |z| = \sqrt{x^2+y^2}$ (modulus). $\theta = \arg(z)$ (argument), $\tan \theta = y/x$. Complex Functions A complex function $f(z)$ maps a complex number $z = x + iy$ to another complex number $w = u + iv$. We write $f(z) = u(x,y) + iv(x,y)$, where $u(x,y)$ is the real part and $v(x,y)$ is the imaginary part of $f(z)$. The goal is to express $u(x,y)$ and $v(x,y)$ in terms of $x$ and $y$. General Approach Substitute $z = x + iy$ into the function $f(z)$. Perform algebraic manipulations (expand, rationalize denominators, use identities). Group terms that do not contain $i$ to find $u(x,y)$. Group terms that are multiplied by $i$ to find $v(x,y)$. Examples of Finding Real and Imaginary Parts 1. Polynomial Function: $f(z) = z^2$ Substitute $z = x+iy$: $f(z) = (x+iy)^2$ Expand: $f(z) = x^2 + 2ixy + (iy)^2 = x^2 + 2ixy - y^2$ Group real and imaginary parts: $f(z) = (x^2 - y^2) + i(2xy)$ Real part: $u(x,y) = x^2 - y^2$ Imaginary part: $v(x,y) = 2xy$ 2. Reciprocal Function: $f(z) = \frac{1}{z}$ Substitute $z = x+iy$: $f(z) = \frac{1}{x+iy}$ Rationalize the denominator by multiplying by the conjugate: $f(z) = \frac{1}{x+iy} \cdot \frac{x-iy}{x-iy} = \frac{x-iy}{x^2 - (iy)^2} = \frac{x-iy}{x^2+y^2}$ Separate into real and imaginary parts: $f(z) = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}$ Real part: $u(x,y) = \frac{x}{x^2+y^2}$ Imaginary part: $v(x,y) = -\frac{y}{x^2+y^2}$ 3. Exponential Function: $f(z) = e^z$ Substitute $z = x+iy$: $f(z) = e^{x+iy}$ Use exponent rule $e^{a+b} = e^a e^b$: $f(z) = e^x e^{iy}$ Use Euler's formula $e^{i\theta} = \cos \theta + i \sin \theta$: $f(z) = e^x (\cos y + i \sin y)$ Distribute $e^x$: $f(z) = e^x \cos y + i e^x \sin y$ Real part: $u(x,y) = e^x \cos y$ Imaginary part: $v(x,y) = e^x \sin y$ 4. Logarithmic Function: $f(z) = \text{Ln}(z)$ (Principal Value) Use polar form $z = re^{i\theta}$, where $r = \sqrt{x^2+y^2}$ and $\theta = \text{atan2}(y,x)$. $f(z) = \text{Ln}(re^{i\theta}) = \ln r + i\theta$. Substitute $r$ and $\theta$ back in terms of $x$ and $y$: $f(z) = \ln(\sqrt{x^2+y^2}) + i \text{atan2}(y,x)$ Simplify $\ln(\sqrt{x^2+y^2}) = \frac{1}{2}\ln(x^2+y^2)$. Real part: $u(x,y) = \frac{1}{2}\ln(x^2+y^2)$ Imaginary part: $v(x,y) = \text{atan2}(y,x)$ (where $-\pi 5. Trigonometric Function: $f(z) = \sin(z)$ Use the definition of $\sin(z)$: $\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$ Substitute $z = x+iy$: $\sin(x+iy) = \frac{e^{i(x+iy)} - e^{-i(x+iy)}}{2i}$ Simplify exponents: $= \frac{e^{ix-y} - e^{-ix+y}}{2i} = \frac{e^{-y}e^{ix} - e^y e^{-ix}}{2i}$ Apply Euler's formula: $= \frac{e^{-y}(\cos x + i\sin x) - e^y(\cos(-x) + i\sin(-x))}{2i}$ Use $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$: $= \frac{e^{-y}(\cos x + i\sin x) - e^y(\cos x - i\sin x)}{2i}$ Expand: $= \frac{e^{-y}\cos x + ie^{-y}\sin x - e^y\cos x + ie^y\sin x}{2i}$ Group terms: $= \frac{(e^{-y}-e^y)\cos x + i(e^{-y}+e^y)\sin x}{2i}$ Multiply by $\frac{-i}{-i}$ to get $i$ out of the denominator: $= \frac{-i[(e^{-y}-e^y)\cos x + i(e^{-y}+e^y)\sin x]}{-2i^2}$ $= \frac{-i(e^{-y}-e^y)\cos x - i^2(e^{-y}+e^y)\sin x}{2}$ $= \frac{(e^{-y}+e^y)\sin x - i(e^{-y}-e^y)\cos x}{2}$ Use definitions of $\cosh y = \frac{e^y+e^{-y}}{2}$ and $\sinh y = \frac{e^y-e^{-y}}{2}$: $= \sin x \cosh y - i(-\sinh y) \cos x$ $= \sin x \cosh y + i \cos x \sinh y$ Real part: $u(x,y) = \sin x \cosh y$ Imaginary part: $v(x,y) = \cos x \sinh y$