eozoischgc

2022-01-12

Give a corect answer for this question how do I get the square root of a complex number? If given a complex number (say 9+4i), how do I calculate its square root?

usaho4w

The square root is not a well defined function on complex numbers. Because of the fundamental theorem of algebra, you will always have two different square roots for a given number. If you want to find out the possible values, the easiest way is probably to go with De Moivres

Timothy Wolff

Here is a direct algebraic answer.
Suppose that $z=c+di$, and we want to find $\sqrt{z}=a+bi$ lying in the first two quadrants. So what are a and b?
Precisely we have
$a=\sqrt{\frac{c+\sqrt{{c}^{2}+{d}^{2}}}{2}}$
and
$b=\frac{d}{|d|}\sqrt{\frac{-c+\sqrt{{c}^{2}+{d}^{2}}}{2}}$
(The factor of $\frac{d}{|d|}$ is used so that b has the same sign as d) To find this, we can use brute force and the quadratic formula. Squaring, we would need to solve
${a}^{2}-{b}^{2}+2abi=c+di$.
This gives two equations and two unknowns (separate into real and imaginary parts), which can then be solved by substitutions and the quadratic formula.

nick1337

You can also do following (technique often advised at school) :
Let's write ${z}^{2}=9+4i$ with $z=a+bi$. The goal is to find z
Thus we have $\left(a+bi{\right)}^{2}=9+4i$ and if you expend we get ${a}^{2}+2abi-{b}^{2}=9+4i$
If you identify the real and imaginary parts, you obtain :
${a}^{2}-{b}^{2}=9$ (1)
and
$2ab=4$ (2)
Now, as ${z}^{2}=9+4i$, the modulus of ${z}^{2}$ and 9+4i are equal so we can write :
${a}^{2}+{b}^{2}={\sqrt{9}}^{2}+{4}^{2}$
${a}^{2}+{b}^{2}=\sqrt{97}$ (3)
Now find a and b with the the equations (1) , (2) and (3) :
$K\left(1\right)+\left(3\right)⇒2{a}^{2}=\sqrt{9+97}$
so
With equation (2) and the previous result you can now find b :
$2ab=4$
$b=2/a$
so $b=2\sqrt{\frac{2}{9+\sqrt{9}7}}orb=-2\sqrt{\frac{2}{9+97}}$
The answer is : $z=\sqrt{\frac{1}{2}\left(9+97\right)}+2i\sqrt{\frac{2}{9+97}}orz=-\sqrt{\frac{1}{2}\sqrt{\left(9+97\right)}}-2i\sqrt{\frac{2}{9+\sqrt{97}}}$

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