# Complex number - roots (1 Viewer)

#### Cujo10

##### Member
I do not get how to solve this question from 2-4. Any help?

#### shashysha

##### Well-Known Member
I think its essentially asking to find the three cube roots of unity of z=1+i. Convert z=1+i into mod-arg form then using De Moivre's Theorem would he what I'd do.

#### Canteen

##### New Member
2.
$\bg_white |z|^{\frac13}=(\sqrt{1^2+1^2})^{\frac13}=\sqrt[6]{2}$
$\bg_white \frac13\textrm{arg}z=\frac13\arctan(1)=\frac\pi{12}$
So the "primary root" will be $\bg_white z=\sqrt[6]{2}\textrm{cis}\frac{\pi}{12}$

3. Note that 1/3arg(z) can also take other values namely:
$\bg_white \frac13\textrm{arg}z=\frac{\pi}{12}+\frac{2\pi}{3}\textrm{ or }\frac{\pi}{12}-\frac{2\pi}{3}=\frac{3\pi}{4}\textrm{ or } -\frac{7\pi}{12}$
So other roots will be:
$\bg_white z=\sqrt[6]{2}\textrm{cis}\frac{3\pi}{4},\ \sqrt[6]{2}\textrm{cis}\left(-\frac{7\pi}{12}\right)$
4. Convert cis roots to cartesian which will be quite laborious in this instance because the arguments aren't nice so compound/double angle formulae for cos and sin will be required.

#### Canteen

##### New Member
I think its essentially asking to find the three cube roots of unity of z=1+i. Convert z=1+i into mod-arg form then using De Moivre's Theorem would he what I'd do.
Technically they are not the "cube roots of unity" but simply the "cube roots".