find cube root of a complex number

Question

I am looking at the following problem and not figuring out how the book's answer is correct. The question is find the cube root of $1+i$. Step 1: convert to Polar: $$r^2 = (1)^2 + (i)^2$$ I see this as $1+(-1)$ which is $0$. The book's answer is that it is $2$ and $r =\sqrt{2}$. Everything else is useless after that point. How does $1$ plus the square of the square root of $-1$ equal anything other than $0$?

Answer 1

Note that for a complex number $$z= a+bi$$ we define $$r=\sqrt {a^2+b^2}$$

For $$z= 1+i = 1+1i $$ we have $$a=b=1$$

Thus $$r=\sqrt {1^2+1^2} = \sqrt 2 $$

Answer 2

If we want to find the complex roots of a number, we use Demoivre's theorem

$$r e^{i\theta (1/n)} = r^{1/n}(e^{i \theta/n})$$

(For brevity's sake, I used $e^{i \theta}$ to represent $\cos \theta + i \sin \theta.$)

To convert an imaginary number $a + bi$ into polar-ready form $r \angle \arg \theta$, we find that $$r = \sqrt {a^2 + b^2} = \sqrt {1 + 1} = \sqrt 2$$ and $$\theta = \arctan \frac {b}{a} = \arctan \frac {1}{1} = \frac {\pi}{4}$$. This now gives us $$\sqrt 2 e^{i \pi /4}$$

Now to find the cube roots...

$$r^{1/3} = 2^{({1/2} \times {1/3})} = 2^{1/6}$$

$$e^{i(\pi /4 \times 1/3)} = e^{i \pi /12}$$

So for the principal cube root of $1 + i$ we have

$$2^{1/6} e^{i \pi/12}$$

To find the other two roots, we have

$$2^{1/6} e^{(i \pi + 2\pi k)/12}, k = 0, 1, 2$$

$k=0$ represents the principal root, so for $k=1$ and $k=2$ we have

$$2^{1/6} e^{i \pi/4}$$

and

$$2^{1/6} e^{5i \pi/12}$$