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.

That's what I would like to know too, it was presented in class like this. However I think it should be transformed into an exponential or logarithmic form. — Brian H Q, Oct 11 '19 at 19:45
For the fundamental system of that DE aren't you really looking for the roots of $r^3-3i=0$? Those are $r_k=\root3\of3e^{i(\pi/6+k2\pi/3)}, k=0,1,2.$ — Jyrki Lahtonen, Oct 12 '19 at 05:11

score 0 · Accepted Answer · answered Oct 11 '19 at 19:45

0

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

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

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

Use Intuition behind euler's formula

answered Oct 11 '19 at 19:45

lab bhattacharjee

274,582

ln 2 / 2 is the same as log(squaredroot( 2))? – Brian H Q Oct 11 '19 at 19:49
@BrianHQ, $$\log(a^m)=m\log a$$ where both logarithms remain defined – lab bhattacharjee Oct 12 '19 at 09:24

score 0 · Answer 2 · answered Oct 11 '19 at 19:47

0

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

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

answered Oct 11 '19 at 19:47

user

154,566

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

2 Answers2