Finding the complex square roots of a complex number without a calculator 
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?
 A: The second equation can be written $ab=2\sqrt{3}$ which gives $b = \frac{2\sqrt{3}}{a}$. If we substitute back into the first equation we get $a^2 - \frac{12}{a^2} = -1 $. Multiplying both sides by $a^2$ gives $a^4 - 12 = - a^2$. This can be written as $a^4 + a^2 - 12 = 0$ which is a quadratic equation solvable for $a^2$.
A: Let$$
z^2=(x+yi)^2=−1+4\sqrt3i,
$$
i.e.$$
(x^2-y^2)+2xyi=−1+4\sqrt3i.
$$
Compare real parts and imaginary parts, $$
\begin{cases}
x^2 - y^2 = -1&\qquad\qquad(1)\\
2xy = 4\sqrt3&\qquad\qquad(2)
\end{cases}
$$
Now, consider the modulus: $|z|^2 =|z^2|$, then
$$x^2 + y^2 = \sqrt{\smash[b]{(-1)^2+(4\sqrt3)^2}} = 7\tag3$$
Solving $(1)$ and $(3)$, we get $x^2 = 3\Rightarrow x = \pm\sqrt3$ and $y^2 = 4\Rightarrow y = \pm2$.
From $(2)$, $x$ and $y$ are of same sign,
$$\begin{cases}
x = \sqrt3\\
y = 2
\end{cases}\text{ or }
\begin{cases}
x = -\sqrt3\\
y = -2
\end{cases}
$$
then$$z = \pm(\sqrt3 + 2i).$$
A: Note that$$z=7\left(-\frac17+\frac{4\sqrt3}7i\right).\tag1$$Now, since $\left(-\frac17\right)^2+\left(\frac{4\sqrt3}7\right)^2=1$, the expression $(1)$ expresses $z$ as $7\bigl(\cos(\alpha)+\sin(\alpha)i\bigr)$, for some $\alpha$. So, a square root of $z$ is $\sqrt7\left(\cos\left(\frac\alpha2\right)+\sin\left(\frac\alpha2\right)i\right)$. Now, note that if $c=\cos\left(\frac\alpha2\right)$ and $s=\sin\left(\frac\alpha2\right)$, then $c^2+s^2=1$ and $c^2-s^2=\cos(\alpha)=-\frac17$. This allows you to compute the square roots of $z$.
A: 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 $(1-2/\sqrt{3}i)\frac{7}{\sqrt{1+4/3}} = \sqrt{7}(\sqrt{3}-2i)$.
A: 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
$$
\pm \left( \sqrt 3 + 2i \right).
$$
A: Observe that $\|z\| = \sqrt{(-1)^2+(4\sqrt3)^2} = \sqrt{49} = 7$. Therefore the root of $z$ will have length $\sqrt 7$, so  $a^2+b^2=7$. Combine this with $a^2-b^2=-1$ to get $a$ and $b$.
