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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?

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  • $\begingroup$ You could notice that $u=a^2$ and $v=-b^2$ satisfy $\begin{cases}u+v=-1\\ uv=-12\\ u\ge 0\\ v\le0\end{cases}$. $\endgroup$
    – user562983
    Commented Oct 5, 2018 at 22:41
  • $\begingroup$ Why is $uv=-12?$ $\endgroup$
    – user376343
    Commented Oct 5, 2018 at 22:53

6 Answers 6

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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 $b$.

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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 $a^2$.

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  • $\begingroup$ 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. $\endgroup$
    – Pegladon
    Commented Oct 5, 2018 at 22:56
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    $\begingroup$ I'm glad that I could help. Don't forget the plus/minus $a=\pm 2i$ and checking your final answer(s). $\endgroup$ Commented Oct 5, 2018 at 23:23
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    $\begingroup$ @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 $a.$ $\endgroup$
    – user376343
    Commented Oct 6, 2018 at 7:49
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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 $$ \pm \left( \sqrt 3 + 2i \right). $$

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  • $\begingroup$ Amazing solution. I haven't thought about using trignonmetry to sole it like this before. +1 $\endgroup$
    – user71207
    Commented Mar 8, 2022 at 1:09
  • $\begingroup$ In general terms this leads to the formula: $\sqrt{a+ib} = \sqrt{r} (a+r+ib)/\lvert a+r+ib \rvert$, where $r=a^2+b^2$. This has a nice graphical interpretation as $a+r+ib$, $a+ib$ and $0$ form an isosceles triangle, and by the Z-angle identity the angles at the base are $\arg(a+ib)/2$ and equal to $\arg(a+r+ib)$. $\endgroup$
    – hife
    Commented Oct 30, 2023 at 11:03
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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$$z = \pm(\sqrt3 + 2i).$$

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    $\begingroup$ Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions. $\endgroup$
    – Théophile
    Commented Oct 5, 2018 at 22:51
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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 $z$.

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

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  • $\begingroup$ Several issues here: 1) The two complex square roots have to be each others negatives. 2) Hence, only one of the solutions of the quadratic equation can be the correct $\tan(\theta)$. 3) The solutions given for the quadratic equation have the wrong sign. 4) $r^2 =7$, so $r=\sqrt{7}$, not 7. $\endgroup$
    – hife
    Commented Oct 30, 2023 at 15:50

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