# 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?

• You could notice that $u=a^2$ and $v=-b^2$ satisfy $\begin{cases}u+v=-1\\ uv=-12\\ u\ge 0\\ v\le0\end{cases}$.
– user562983
Oct 5, 2018 at 22:41
• Why is $uv=-12?$ Oct 5, 2018 at 22:53

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 $$b$$.

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 $$a^2$$.

• Thanks, I managed to get up to $a^4 + a^2 - 12 = 0$ on my own before asking this question, but couldn't figure out how to solve for $a$ without a calculator. Using your suggestion I found $(a^2 + 4)(a^2 - 3) = 0$ therefore $a = 2i \text{ and } \pm \sqrt{3}$, thanks for the help. Oct 5, 2018 at 22:56
• I'm glad that I could help. Don't forget the plus/minus $a=\pm 2i$ and checking your final answer(s). Oct 5, 2018 at 23:23
• @Pegladon, in this method $a,b$ are considered reals. Moreover, any non-zero complex number has exactly two square roots. Thus $\pm 2i$ are not convenient for $a.$ Oct 6, 2018 at 7:49

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 $$(1-2/\sqrt{3}i)\frac{7}{\sqrt{1+4/3}} = \sqrt{7}(\sqrt{3}-2i)$$.

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 $$\pm \left( \sqrt 3 + 2i \right).$$

• Amazing solution. I haven't thought about using trignonmetry to sole it like this before. +1 Mar 8 at 1:09

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$$z = \pm(\sqrt3 + 2i).$$

• Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions. Oct 5, 2018 at 22:51

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 $$z$$.