0
$\begingroup$

With $ω=se^{iφ}$, where $s≥0$ and φ∈ , solve the equation $z^n=ω$ in C where n is a natural number. How many solutions are there?

I have the solutions to this problem however, do not understand its steps.

I do not understand how $z=s^\frac 1n e^{(i\frac \phi n+\frac{2kpi}n)}$ where $k \in Z$.

Where is the $2k\pi$ coming from?

Any verification will help thank you

1 Answers 1

0

Using this, $e^{ix}=\cos x+i\sin x$

So, $e^{2k\pi i}=\cos {2k\pi}+i\sin {2k\pi}=1$ if $k$ is any integer.

$ω=se^{i\phi}$

$z^n=w=se^{i\phi}$

This is an $n-$degree equation in $z,$ so has $n$ solutions.

$z^n=se^{i\phi}\cdot1=se^{i\phi}e^{2k\pi i}=se^{i(\phi+2k\pi)}$

$z=s^{\frac 1 n}e^{\frac{i(\phi+2k\pi)}n} $ where $0\le k , more generally any $n$ in-congruent values of $k$ will give us essentially the same set of $n$ distinct solutions, find the explanation here .