# Roots of complex numbers - true/false

Here's another one of the problems I'm having trouble with. Below are two statements and the book is asking me to tell for each whether it is true or false:

1. All the solutions to the equation $\ \ z^3=i\ \$ are:

$z_1=-i\ ,\ z_2=\frac{\sqrt3}{2}+\frac{i}{2}\ ,\ z_3=-\frac{\sqrt3}{2}+\frac{i}{2}$

1. All the solutions to the equation $\ \ z^2=i-\sqrt3\ \$ are:

$z_1=\sqrt2\left( \cos\frac{\pi}{3}+i\sin\frac{\pi}{3} \right)\ ,\ z_2=\sqrt2\left( \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} \right)$

I know that the three answers listed for the first one are indeed roots of $i$ but I'm not so sure about the second statement.

• $z^2+(\sqrt{3}-i)$ has exactly $2$ complex roots by the Fundamental Theorem of Algebra. Clearly $z_1\neq z_2$, so all you would have to do is to actually compute $z_1^2$ and $z_2^2$ and verify if they, indeed, satisfy the equation. – Guest Dec 18 '15 at 20:16
• For the first: $i^3 = -i \neq i$. – Théophile Dec 18 '15 at 20:29
• Oh I think I copied the first part wrong - $z_1$ should be $-i$. Sorry about that. – Or Bairey-Sehayek Dec 18 '15 at 20:32

## 1 Answer

$$z^2=i-\sqrt{3}\Longleftrightarrow$$ $$z^2=|i-\sqrt{3}|e^{\arg(i-\sqrt{3})i}\Longleftrightarrow$$ $$z^2=2e^{\frac{5\pi i}{6}}\Longleftrightarrow$$ $$z=\left(2e^{\left(2\pi k+\frac{5\pi}{6}\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow$$ $$z=\sqrt{2}e^{\frac{1}{2}\left(2\pi k+\frac{5\pi}{6}\right)i}$$

With $k\in\mathbb{Z}$ and $k:0-1$

So the solutions are:

$$z_0=\sqrt{2}e^{\frac{1}{2}\left(2\pi\cdot0+\frac{5\pi}{6}\right)i}=\sqrt{2}e^{\frac{5\pi i}{12}}=\sqrt{2}\left(\cos\left(\frac{5\pi}{12}\right)+\sin\left(\frac{5\pi}{12}\right)i\right)$$ $$z_1=\sqrt{2}e^{\frac{1}{2}\left(2\pi\cdot1+\frac{5\pi}{6}\right)i}=\sqrt{2}e^{-\frac{7\pi i}{12}}=\sqrt{2}\left(\cos\left(-\frac{7\pi}{12}\right)+\sin\left(-\frac{7\pi}{12}\right)i\right)$$

Notice with $a,b\in\mathbb{R}$: $$a+bi=|a+bi|e^{\arg(a+bi)i}=|a+bi|\left(\cos(\arg(a+bi))+\sin(\arg(a+bi))i\right)$$

With $e$ is the base of the natural logarithm.

• I think what you used here is polar notation and we never learned that in class. What does the $e$ mean here? If I'm understanding correctly, $e^{\frac{5\pi i}{12}}$ means $\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}$ - is that correct? – Or Bairey-Sehayek Dec 18 '15 at 20:20
• @OrBairey-Sehayek Look to my edit, I hope it is understandable now – Jan Eerland Dec 18 '15 at 20:24
• Ah okay, thanks. – Or Bairey-Sehayek Dec 18 '15 at 20:25
• @OrBairey-Sehayek You're more than welcome! – Jan Eerland Dec 18 '15 at 20:26