Square root of $i$ using polar form

Question

Suppose I want to find the square root of $i$ using polar form. We have:

$z^2 = 0+i = \cos{\dfrac{\pi}{2}} + i \sin\dfrac{\pi}{2} = e^{i\dfrac{\pi}{2}}$

Then $z = e ^{i\dfrac{\pi + 2k\pi}{4}}$

$k=0 \Rightarrow z=e^{\dfrac{\pi}{4}}$

$k=1 \Rightarrow z=e^{\dfrac{3\pi}{4}}$

However, we can see that these are not complements to each other. What am I doing wrong here?

Any help would be appreciated.

Thanks.

Answer 1

You have

$z = e ^{i\dfrac{\pi + 2k\pi}{4}}$

The i$\frac{\pi}{4}$ term is right but you needed to divide the periodicity term $i2k\pi$ by $2$ not $4$. Dividing the periodicity term by $4$ corresponds to a fourth root. Thus, properly,

$z = e ^{i(\dfrac{\pi}{4} + \dfrac{2k\pi}{2})}$

Now put $k=1$ and see what is gotten.

Answer 2

I suppose that whe you wrote “complements”, you actually meant “conjugate”. Anyway, the square roots of $\exp(i\theta)$ are the numbers $\exp\left(\frac{i\theta+2k\pi i}2\right)=\exp\left(\frac{i\theta}2+k\pi i\right)$ ($k\in\mathbb N$). So, if we put $\theta=\frac\pi2$, what we get are the numbers $\exp\left(\frac{i\pi}4\right)$ and $\exp\left(\frac{i\pi}4+\pi i\right)$, which are conjugate of each other.