complex exponential equation $e^{jz\pi}+j=0$

Question

Which are the solutions of this equation?

$e^{jz\pi}+j=0$

i was using the formula

$\cos(...)+j\sin(...)$ but it gives wrong results... I can't understand how to get it step by step

Answer 1

Recall that

$$-j=e^{j\left(\frac{3\pi}2+2k\pi\right)}$$

therefore

$$e^{jz\pi}+j=0 \iff e^{jz\pi}=-j=e^{j\left(\frac{3\pi}2+2k\pi\right)}$$

that is

$$z\pi=\frac{3\pi}2+2k\pi \implies z=\frac{3}2+2k \quad k\in \mathbb{Z}$$

Answer 2

You have to solve $e^{jz\pi}=-j$. Now $-j=e^{-j\pi/2}$ and $e^a=e^b$ if and only if $a-b=2k\pi j$ for an integer $k$. Can you take it from there?

Answer 3

You can use the fact that $e^a=e^b$ if and only if $a=b+2\pi k i$ for integral $k$.

Put your equation in this form as follows:

$$e^{iz\pi} + i=0$$ $$e^{iz\pi}=-i$$ $$e^{iz\pi}=e^{-\pi i/2}\tag{since $-i$ can be written as $e^{-\pi i/2}$}$$ So this is true if and only if

$$iz\pi = -\pi i/2 + 2\pi ki = \pi i(2k - \tfrac12)$$ $$\boxed{z = 2k-\tfrac12}$$ for integral $k$. These are the values $$\left\{\ldots, -\frac92, -\frac52,-\frac12,\frac32,\frac72,\frac{11}2,\ldots \right\}$$