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How do I get the square root of a complex number?

I know that the answer to $\sqrt{\dfrac{-3}{4} - i}$ is $\dfrac12 - i$. But how do I calculate it mathematically if I don't have access to a calculator?

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marked as duplicate by Hans Lundmark, t.b., J. M., Zev Chonoles Dec 17 '11 at 7:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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If $\sqrt{a+bi}=c+di$, then $a+bi=c^2-d^2+i(2cd)$. You now have two equations in two unknowns, corresponding to the real and imaginary components. –  J. M. Dec 16 '11 at 16:24
    
Note that that there are always two square roots of any complex number different than zero (if you multiply a number by $-1$ and compute square of it, you get the same). It is also true for real numbers but people agreed to interpret $\sqrt{r}$ as the positive root of r (when they use real numbers only). –  savick01 Dec 16 '11 at 21:05

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up vote 4 down vote accepted

One of the standard strategies (the other strategy is to do what JM suggested in the comment to the qn) is to complete the square and use the fact that $i^2 = -1$.

$$\sqrt{\frac{-3}{4}-i}$$

Add and subtract 1 to get:

$$\sqrt{\frac{-3}{4}+1-i-1}$$

Use $i^2 = -1$ to get:

$$\sqrt{\frac{-3}{4}+1-i+i^2}$$

Simplify $\frac{-3}{4}+1$ to get:

$$\sqrt{\frac{1}{4}-i+i^2}$$

Rewrite $-i$ as $-2 \frac{1}{2}i$ to get:

$$\sqrt{\frac{1}{2^2}-2 \frac{1}{2}i+i^2}$$

Complete the square to get:

$$\sqrt{(\frac{1}{2}-i)^2}$$

Get rid of the square root to get:

$$\frac{1}{2}-i$$

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Thank you a lot –  Dimme Dec 16 '11 at 16:36

Another method is polar coordinates. $|-3/4-i| = 5/4$ and
$$ -\frac{3}{4} - i = \frac{5}{4}\;\exp(i\theta) $$
where $\cos(\theta) = -3/5 $ and $\sin(\theta) = -4/5 $. Use half-angle formulas to get $\cos(\theta/2)$ and $\sin(\theta/2)$, actually two choices, then your answer is
$$ \sqrt{\frac{5}{4}}\;\exp\left(\frac{i\theta}{2}\right) $$

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For $a \ne 0$, there are two numbers $z$ such that $z^2=a$.

We look at the given example, using only basic tools. We want to solve the equation $$z^2=-\frac{3}{4}-i.$$ Because of a general discomfort with negative numbers, we look at the equivalent equation $$z^2=-\frac{1}{4}(3+4i).$$ In order to deal with simpler numbers, we first consider the equation $$w^2=3+4i.$$ Let $w=a+ib$. Then $w^2=(a^2-b^2)+2ab i$. So we want to solve the system of two equations $$a^2-b^2=3, \qquad 2ab=4.$$ The solutions can be found by inspection. However, we continue, in order to show how to proceed when the numbers are less simple. Rewrite the second equation as $b=2/a$. Substitute in the first equation. We get $$a^2-\left(\frac{2}{a}\right)^2=3,$$ which after some simplification becomes $$a^4-3a^2-4=0.$$ This is a quadratic equation in $a^2$. By using the Quadratic Formula, we find that the roots are $a^2=4$ and $a^2=-1$. The second equation has no real solution, so $a=\pm 2$. We get the two solutions $a=2$, $b=1$ and $a=-2$, $b=-1$.

Thus $w=2+i$ or $w=-(2+i)$. So find $z$, multiply these two values of $w$ by $\dfrac{i}{2}$.

Another way: Any complex number $z^2$ can be written as $r(\cos \theta+i\sin\theta)$ where $r$ is non-negative. Then $$z^2=r^2[(\cos^2\theta)+i(2\cos \theta\sin\theta)].$$ We can rewrite this as $r^2(\cos(2\theta)+i\sin(2\theta)$. We want $z^2=-\frac{3}{4}-i$. The norm of $-\frac{3}{4}-i$ is the square root of $(-3/4)^2+(-1)^2$, so $r^2=(9/16)+1=25/16$ and $r=5/4$. We know that $\cos(2\theta)=(4/5)(-3/4)$ and $\sin(2\theta)=(4/5)(-1)$.

Simplify. We get $\cos(2\theta)=-\frac{3}{5}$ and $\sin(2\theta)=-\frac{4}{5}$. Now we could proceed by calculator, finding $2\theta$, then $\theta$, then $\cos\theta$ and $\sin\theta$. Or else we can proceed algebraically, using the fact that $\cos^2\theta-\sin^2\theta=-\frac{3}{5}$ and $2\sin\theta\cos\theta=-\frac{4}{5}$. If we do this, the rest is much like the first solution. We have $\sin\theta=(-4/5)/(2\cos\theta)$. Substitute in $\cos^2\theta-\sin^2\theta=-\frac{4}{5}$. After simplifying, we obtain a quadratic equation in $\cos^2\theta$, and the rest is routine. There will be two values of $\cos\theta+i\sin\theta$ that work, and they will be the negatives of each other, and the roots are $\pm\sqrt{r}(\cos\theta+i\sin\theta)$.

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