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I know that the square root of a number x, expressed as $\displaystyle\sqrt{x}$, is the number y such that $y^2$ equals x. But is there any simple way to calculate this with complex numbers? How?


marked as duplicate by MJD, Henry T. Horton, Chris Godsil, rschwieb, Micah Jun 4 '13 at 18:19

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    $\begingroup$ Do you know about the polar representation of complex numbers? $\endgroup$ – Milind Jun 4 '13 at 17:08
  • $\begingroup$ @MJD No, I'm talking complex here, not only imaginary. And could you maybe give a 3D plot? Thanks guys! $\endgroup$ – Anonymous Pi Jun 4 '13 at 17:09
  • $\begingroup$ @AnonymousPi Please read the answers I linked to; they answer your question. Or try square root of complex number $\endgroup$ – MJD Jun 4 '13 at 17:10
  • $\begingroup$ Hey! Why the downvote? I'm not that advanced in math as you guys! $\endgroup$ – Anonymous Pi Jun 4 '13 at 17:11
  • $\begingroup$ My guess is that the downvote is because it is a duplicate of the question that MJD linked. (Although the question is posed as $\sqrt{i}$, the answers are more generally applicable to complex numbers) It is not a reflection of the quality of your question, but is probably instead a signal to others that it should be closed as a duplicate. $\endgroup$ – Eric Stucky Jun 4 '13 at 17:17

If you represent the complex number $z$ in polar form, i.e.

$$z=re^{i\theta} = r(\cos \theta+i\sin\theta)$$

where $r>0, \theta \in [0, 2\pi)$.

Then the square roots of $z$ are

$$\sqrt z = \pm \sqrt re^{i\theta/2}$$

In general the $k$ $k$th roots of $z$ are $\sqrt[k]r\exp\left(i\times\left(\frac{2\pi j}{k}+\frac\theta k\right)\right)$ for $j=0,1,2,...,k-1.$


Sure. By the fundamental theorem of algebra, the complex numbers $\mathbb{C}$ are an algebraically closed field. Informally, what this means is that any polynomial equation with coefficients in $\mathbb{C}$ has roots in $\mathbb{C}$. Say, then, that you have a complex number $a+bi$ that you want to find the square root of. What you really want to do is find the solutions to the equation $x^{2} - (a+bi) = 0$ - you can see, for example, that $i$ is the solution to the equation $x^{2} - (-1) = 0$.

The fundamental theorem of algebra tells us that our roots will be in $\mathbb{C}$, i.e., will be expressible as $c+di$ for some $c, d \in \mathbb{R}$.

Plugging in, then, and equating real and imaginary parts, we see:

$$(c+di)^{2} = a+bi$$

$$\implies c^{2} - d^{2} = a;~~ 2cd = b$$

For some fixed $a, b \in \mathbb{R}$, we can solve the equation thusly.

Assuming $c \neq 0$, which certainly must be true in the case that $b \neq 0$ (in which case the answer is obvious), we note that:

$$d = \frac{b}{2c}$$

$$\implies c^{2} - (\frac{b}{2c})^{2} = a;$$ With some algebra, we see:

$$\implies c^{4} - 4ac^{2} - b^{2} = 0$$

Let $k = c^{2}$. Then this easily transforms to a quadratic which we can solve:

$$k^{2} - 4ak -b^{2} = 0;$$ Using the quadratic formula, then, we get:

$$k = \frac{4a \pm \sqrt{16a^{2} + 4b^{2}}}{2} = 2a \pm \sqrt{4a^{2} + b^{2}}$$

Note: the discriminant must be nonnegative, so $k$ is strictly real. This means that we don't have to worry about $c$ being the square root of a complex number, which would defeat the purpose of this method.

Then, $$c = \pm\sqrt{k} = \pm \sqrt{2a \pm \sqrt{4a^{2} + b^{2}}}$$ and $$d = \frac{b}{2c}$$

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    $\begingroup$ Why stop in the middle; why not solve the equations? From the first equation you get $c = \pm\sqrt{a+d^2}$… $\endgroup$ – MJD Jun 4 '13 at 17:17
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    $\begingroup$ @MJD, I've updated my solution accordingly. $\endgroup$ – Alex Wertheim Jun 4 '13 at 17:31

Write $(x+iy)^2=a+ib$, thus $x^2-y^2=a$ and $2xy=b$.

Now, $-x^2y^2=-b^2/4$, so $x^2$ and $-y^2$ are the solutions of $t^2-at-b^2/4=0$.

Then $\Delta=a^2+b^2$, and $t=\frac{a \pm \sqrt{a^2+b^2}}{2}$.

Obviously, since $x^2 \geq 0$ and $-y^2 \leq 0$, we have

$$x^2 = \frac{\sqrt{a^2+b^2}+a}{2}$$ $$y^2 = \frac{\sqrt{a^2+b^2}-a}{2}$$

Then, using the fact that $xy$ has the same sign than $b$, we have the two square roots of $a+ib$,

$$x+iy = \pm \left( \sqrt{\frac{\sqrt{a^2+b^2}+a}{2}} + i \ \mathrm{sign}(b) \sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}\right)$$


If you write the number $a+bi$ as $re^{i\phi}$ you can deduce: $\sqrt{a+bi}=\sqrt{re^{i\phi}}=e^{\pm \frac{i\phi}{2}}\sqrt{r}$

$r=\sqrt{a^2+b^2}$, $\phi=\arctan{\frac{b}{a}}$

  • $\begingroup$ This seems incomplete. You can expand $e^{ix}$ as $\cos x + i\sin x$, and when $x$ is $\arctan \psi$ you can eliminate the trigonometry entirely. $\endgroup$ – MJD Jun 4 '13 at 17:14
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    $\begingroup$ Where's the other root? $\endgroup$ – egreg Jun 4 '13 at 17:22
  • $\begingroup$ i forgot to add the $\pm$... $\endgroup$ – V-X Jun 4 '13 at 17:29
  • $\begingroup$ I don't like very much to see $\arg (a+ib)$ written as $\arctan \frac{b}{a}$... $\endgroup$ – Jean-Claude Arbaut Jun 4 '13 at 17:30

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