How can I solve $z^{-i} = 1+i$

Question

Basically I need to solve $z^{-i} = 1+i$

But the $-i$ is throwing me off.

As well, talking about differential equations how can I get the fundamental system of solutions of $y'''-3iy=0 $. I know it is a homogenous equation but the $i$ following the $y$ is throwing me off.

Thanks in advance.

Answer 1

$$-i\ln z=\ln(1+i)=\ln(\sqrt2e^{i\pi/4})=\dfrac{\ln2}2+i\left(2n\pi+\dfrac\pi4\right)$$

$$\ln z=i\cdot\dfrac{\ln2}2-\left(2n\pi+\dfrac\pi4\right)$$

$$z=e^{i\cdot\dfrac{\ln2}2-\left(2n\pi+\dfrac\pi4\right)}=e^{-\left(2n\pi+\dfrac\pi4\right)}\cdot e^{i\cdot\dfrac{\ln2}2}$$

Use Intuition behind euler's formula

Answer 2

By $z=re^{i\theta}$ we have

$$z^{-i}=1+i \implies r^{-i}e^{\theta}=\sqrt 2\cdot e^{i(\frac{\pi}4+2k\pi)}$$

that is

• $e^{\theta}=\sqrt 2$
• $r=e^{-(\frac{\pi}4+2k\pi)}$