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How can one find the square roots of a complex number (say $-7+13i$, or more generally $a+bi$) using de Moivre's formula?

I've begun a course on complex variables recently and one of the first things we've covered is complex numbers in polar form, where de Moivre's formula emerges. But in this, I know it can be used to find the two square roots of a complex number, but I'm really not sure how. I feel I might be getting lost in the trigonometry.

Sayan Dutta
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  • It will be better for this case to use the algebraic form of complex numbers. Consider $x + i y$ a square root of $-7 + 13 i$ then : $$-7 + 13 i = x^2 - y^2 + 2 i x y$$ and : $$x^2 + y^2 = |x + i y|^2 = \sqrt{7^2 + 13^2}$$ Use the system : $$\left{\begin{array}{lcl} x^2 - y^2 & = & -7 \ x^2 + y^2 & = & 7^2 + 13^2 \end{array}\right.$$ to determiner $x^2$ and $y^2$ and the equation $2 x y = 13$ for the sign of $x$ and $y$. – Essaidi Jan 30 '22 at 16:26
  • A comment rather than an answer as this is surely a duplicate and has been asked many times but Step 1) (the hard step) Convert $a+bi$ to $r e^{\theta i}=re^{\theta i+2k\pi i}$ form. then Step 2) (the easy step) $(re^{\theta i+2k\pi i})^{\frac 12}=r^{\frac 12} e^{\frac \theta 2i +k\pi} =\sqrt r e^{\frac \theta 2i},\sqrt r e^{(\frac \theta 2 +\pi)i}$. .... Now step 1.... to be continued..... – fleablood Jan 30 '22 at 16:51
  • ....$z=a+bi =|z|( \frac a{|z|} + i\frac b{|z|})=|z|(\frac a{\sqrt{a^2 + b^2}} + i\frac b{\sqrt{a^2+b^2}})$. As $(\frac {a}{\sqrt{|z|}})^2 + (\frac {b}{|z|})^2 = \frac {a^2+b^2}{a^2 + b^2}= 1$ then $(\frac a{|z|},\frac b{|z|}$ is on the unit circle and there is a unique $\theta$ so that $\frac a{|z|}=\cos \theta$ and $\frac b{|z|}=\sin\theta$. To find that that $\theta$ note $\tan \theta=\frac {\frac{b}{|z|}}{\frac {a}{|z|}}=\frac ba$ so $\theta = \arctan \frac ba$. So $z=\sqrt{a^2+b^2}(\cos \arcsin \frac ba+i\sin \arcsin \frac ba)=|z|e^{\arcsin \frac{Re(z)}{Im(z)}i}$. – fleablood Jan 30 '22 at 16:57

3 Answers3

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Write the complex number $z$ as $r(\cos\theta+i\sin\theta)$ and use De'Moivre's formula to claim that $$\sqrt z=r^{\frac 12}\left(\cos \left(\frac \theta 2\right)+i\sin \left(\frac \theta 2\right)\right)$$

Sayan Dutta
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Well, we know that:

$$\text{z}=\Re\left(\text{z}\right)+\Im\left(\text{z}\right)i=\left|\text{z}\right|\exp\left(\left(\text{Arg}\left(\text{z}\right)+2\pi\text{k}\right)i\right)\tag1$$

Where $\text{Arg}\left(\text{z}\right)$ is the principle value of the argument, $\text{k}\in\mathbb{Z}$ and $\left|\text{z}\right|=\sqrt{\Re\left(\text{z}\right)^2+\Im\left(\text{z}\right)^2}$.

Now, taking the square root gives:

\begin{equation} \begin{split} \sqrt{\text{z}}&=\text{z}^\frac{1}{2}\\ \\ &=\left(\left|\text{z}\right|\exp\left(\left(\text{Arg}\left(\text{z}\right)+2\pi\text{k}\right)i\right)\right)^\frac{1}{2}\\ \\ &=\left|\text{z}\right|^\frac{1}{2}\cdot\exp\left(\frac{1}{2}\cdot\left(\text{Arg}\left(\text{z}\right)+2\pi\text{k}\right)i\right)\\ \\ &=\sqrt{\left|\text{z}\right|}\cdot\exp\left(\left(\frac{\text{Arg}\left(\text{z}\right)}{2}+\pi\text{k}\right)i\right)\\ \\ &=\sqrt{\left|\text{z}\right|}\cos\left(\frac{\text{Arg}\left(\text{z}\right)}{2}+\pi\text{k}\right)+\sqrt{\left|\text{z}\right|}\sin\left(\frac{\text{Arg}\left(\text{z}\right)}{2}+\pi\text{k}\right)i \end{split}\tag2 \end{equation}

Jan Eerland
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First you should find $r$ and $\theta$ like below. Since $r^2=a^2+b^2$

$$\sqrt{a^2+b^2}=\sqrt{r^2}=\begin{cases} r,~~~r>0\\ -r,~~~r<0 \end{cases}$$

But since $r$ is positive,

$$r=\sqrt{a^2+b^2}$$

Also $\tan\theta=\frac ba$, So

$$\operatorname{Arctan}\frac ba=\operatorname{Arctan}(\tan\theta)=\begin{cases} \theta,~~~\text{first quadrant}\\ \theta-\pi,~~~\text{second and third quadrants}\\ \theta-2\pi,~~~\text{fourth quadrant} \end{cases}$$

and

$$\theta=\begin{cases} \operatorname{Arctan}\frac ba,~~~(a,b>0)\\ \pi+\operatorname{Arctan}\frac ba,~~~(a<0)\\ 2\pi+\operatorname{Arctan}\frac ba,~~~(a>0,b<0) \end{cases}$$

Then

$$\begin{align} \sqrt z&=\sqrt r\left(\cos\left(\frac{\theta+2k\pi}2\right)+i\sin\left(\frac{\theta+2k\pi}2\right)\right), k=0,1\\ &=\pm\sqrt{r}\left(\cos\left(\frac{\theta}2\right)+i\sin\left(\frac{\theta}2\right)\right) \end{align}$$

Sidvhid Hsinynjad
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