How do I get the square root of a complex number? If I'm given a complex number (say $9 + 4i$), how do I calculate its square root?

You can also do following (technique often advised at school) :


*

*Let's write $z² = 9 + 4i$ with $z = a + bi$.  The goal is to find $z$
Thus we have  $(a + bi)² = 9 + 4i$ and if you expend we get $a²+ 2abi - b² = 9 + 4i$
If you identify the real and imaginary parts, you obtain :
$a²-b² = 9$   (1) 
and 
$2ab= 4$   (2)

*Now, as $z² = 9 + 4i$, the modulus of $z²$ and $9 + 4i$ are equal so we can write :
$a²+b² = \sqrt{9²+4²}$
$a²+b² = \sqrt{97}$   (3) 

*Now find $a$ and $b$ with the the equations (1) , (2) and (3) : 
(1) + (3) $\Leftrightarrow 2a² = 9+\sqrt{97} $ 
so $a = \sqrt{\frac{1}{2}(9+\sqrt{97})} $  or  $a = - \sqrt{\frac{1}{2}(9+\sqrt{97})} $
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{97}}} $  or  $b = - 2\sqrt{\frac{2}{9+\sqrt{97}}} $
The answer is : $z = \sqrt{\frac{1}{2}(9+\sqrt{97})} + 2i\sqrt{\frac{2}{9+\sqrt{97}}}  $ or $z = - \sqrt{\frac{1}{2}(9+\sqrt{97})} - 2i\sqrt{\frac{2}{9+\sqrt{97}}}  $
