$z^5=-2$ Complex number solutions

Question

Find all complex number solutions to the equation $$z^5=-2$$

I'm a little lost in using De Moivre's theorem and Euler's formula.

Answer 1

Euler's famous identity tells us that $e^{i(2k+1)\pi} = -1$, so $$2e^{i\pi(2k+1)} = -2 = z^5$$ Now take the fifth root of each side to solve for $z$ to get $$z = 2^{\frac{1}{5}}e^\frac{i(2k+1)\pi}{5}$$ You'll get five solutions by plugging in $k = 0,1,2,3,4$ into the complex exponent. Use the identity $e^{ix} = \cos(x) + i\sin(x)$ to convert your answer back to Cartesian coordinates, if you need to.

Answer 2

If $z^n = a$ where $a\neq 0$ then $$z = \sqrt[n]{|a|}e^{i\left(\frac{\arg(a)}{n}+p\frac{2\pi}{n}\right)}$$ where $p\in\{0,...,n-1\}$, which gives $n$ solutions. Then use that $$e^{i\theta} = \cos\theta+i\sin\theta $$ Does this help?

Answer 3

To visualize roots on the complex Argand Diagram $ ( x + i y)$ rotate vector between $(0,0)$ and $(-2^{1/5},0)$ five times through an angle $ 2 \pi/5 $ around $(0,0).$