TITLE: Finding the complex square roots of a complex number without a calculator
QUESTION [7 upvotes]: The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$

The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4\sqrt{3})i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?

REPLY [3 votes]: One way is to write $z=r^2 e^{2\theta}$ and roots will be $re^\theta$ and $re^{\pi -\theta}$. 
From $z =-1+4\sqrt{3}i$, we obtain $r=7$ and $\tan{2\theta} = \frac{2\tan\theta}{1-\tan^2\theta} = -4\sqrt{3}$. Second expression gives you a quadratic equation, $2\sqrt{3}\tan^2\theta -2\sqrt{3} + \tan\theta =0$. 
Roots of the above quadratic equation are $\tan\theta= \sqrt{3}/2,-2/\sqrt{3}$ which form $\tan\theta$ and $\tan(\pi-\theta)$. 
Hence, square roots of $z$ are $(1+\sqrt{3}/2i)\frac{7}{\sqrt{1+3/4}} = \sqrt{7}(2+\sqrt{3}i)$ and $(1-2/\sqrt{3}i)\frac{7}{\sqrt{1+4/3}} = \sqrt{7}(\sqrt{3}-2i)$.

REPLY [3 votes]: That the square roots are $\pm(\sqrt 3 + 2i)$ can be seen by elementary algebra and trigonometry as follows.
\begin{align}
& \left|-1 + i4\sqrt 3\right| = \sqrt{(-1)^2 + (4\sqrt 3)^2 } = 7. \\[10pt]
\text{Therefore } & -1+i4\sqrt 3 = 7(\cos\varphi + i\sin\varphi). \\[10pt]
\text{Therefore } & \pm\sqrt{-1+i4\sqrt 3} = \pm\sqrt 7 \left( \cos \frac \varphi 2 + i \sin\frac\varphi 2 \right).
\end{align}
Notice that
$$
\sin \varphi = \frac{4\sqrt 3} 7 \quad \text{and} \quad \cos\varphi = \frac{-1} 7
$$
and recall that
\begin{align}
\tan\frac\varphi 2 & = \frac{\sin\varphi}{1+\cos\varphi} \\[12pt]
\text{so we have }\tan\frac\varphi 2 & = \frac{4\sqrt 3}{7-1} = \frac 2 {\sqrt 3}. \\[10pt]
\text{Therefore } \sin\frac\varphi2 & = \frac 2 {\sqrt 7} \quad \text{and} \quad \cos\frac\varphi2 = \frac{\sqrt3}{\sqrt7}.
\end{align}
Thus the desired square roots are
$$
\pm \left( \sqrt 3 + 2i \right).
$$