Square root of an algebraic form of a complex number

Question

I have $$ \sqrt{1+i\sqrt{3}}=\sqrt{2}\left[\cos\left(\frac{\frac{\pi}{3}+2k\pi}{2}\right)+i\sin\left(\frac{\frac{\pi}{3}+2k\pi}{2}\right)\right]=\sqrt{2}\left(\pm \frac{\sqrt{3}}{2}\pm\frac{1}{2}i\right). $$

My question is: why do we add $2k\pi$ to the argument of the trigonometric functions?

Answer 1

If we view $1 + \sqrt 3i$ as the point $(1,\sqrt 3)$ in a plane it is a distance of $\sqrt {1 + \sqrt{3}^2} =2$ from the origin. It is at a specific angle to thee $x$-axis. It is the angle whose sine is $\frac {\sqrt 3}2$ and whose cosine is $\frac 12$. That angle is $\frac {\pi}3$.

So we can write $1+\sqrt 3i$ as $2(\cos \frac {\pi}3 + i \sin \frac {\pi}3)$.

All non-zero complex numbers can be written as some $r(\cos \theta + i \sin \theta)$. The advantage is that if you multiply to complex numbers represented this way you get $r(\cos\theta + i \sin \theta)\cdot s(\cos \phi + i\sin \phi)= rs (\cos (\theta +\phi)+ i \sin(\theta + \phi))$. This makes multiplication and division easy; just multiple their magnitudes and add their angles.

So to solve $[r(\cos \theta + i \sin \theta)]^2 = 1 + i\sqrt 3= 2(\cos \frac {\pi}3 + i \sin \frac {\pi}3)$ we just have to solve

$r^2 (\cos 2\theta + i \sin 2\theta) =2(\cos \frac {\pi}3 + i \sin \frac {\pi}3)$. Or in other words $r^2 = 2$ and $2\theta = \frac {\pi}3$.

As $r$ is a distance, it must be positive so $r = \sqrt 2$. But we have to be careful about $\theta$.

If $\theta > \pi$ then $2\theta > 2\pi$ and we have wrapped around the circle. We must consider that not only might $2\theta = \frac {\pi}3$ but $2\theta$ might equal $\frac {\pi}3 + 2\pi$.

So $\theta = \frac {\pi}6$ or $\theta = \frac {7\pi} 6$.

So $\sqrt{1 + i\sqrt 3} = \sqrt 2(\cos \frac {\pi}6 + i\sin \frac {\pi}6)$ or $=\sqrt 2(\cos \frac {7\pi}6 + i\sin \frac {7\pi}6)$.

Or $\sqrt{1+i\sqrt 3} = \sqrt 2(\frac {\sqrt 3}2 + i\frac 12)$ or $= \sqrt 2(-\frac {\sqrt 3}2 - i\frac 12)$

To answer you question why do we and $2k\pi$. I hope it's clear in that it's to consider the cases in solving that you might have gone "round the circle" So $\sin|\cos (\theta) = \sin|\cos (\theta + 2k\pi)$ this will allow you to solve if $\cos(n\theta) + i\sin(n\theta) = \cos(\phi) + i\sin(\phi)$.

To solve $\cos(n\theta) + i\sin(n\theta) = \cos(\phi) + i\sin(\phi)=\cos(\phi+2k\pi) + i\sin(\phi+2k\pi)$ so

$n\theta = \phi + 2k\pi$ so $\theta = \frac {\phi}n + \frac {2k}n\pi$. We can't overlook the solutions that will "take us round the circle"