In simplifying the formula that I've derived for finding the square root of a complex number to the standard formula.

Question

So by easy means, I derived

$\sqrt{a+ib} = \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$

But then I checked for the actual formula it is this;

$\sqrt{a+ib} = \sqrt{\frac{\sqrt{a^2 + b^2}+a}{2}}± i(\sqrt{\frac{\sqrt{a^2 + b^2}-a}{2}})$

So how do you simplify that to this? Btw here is the derivation;

Let $\sqrt{a+ib} = x+iy$

$a+ib=x^2 - y^2 + i2xy$

We know, $(x^2-y^2)^2 + (2xy)^2 = x^2 + y^2$

or, $x^2 + y^2 = a^2 + b^2$

Hence we get $x = \sqrt{\frac{a+a^2+b^2}{2}}$ and $y=\frac{b}{\sqrt{2}\sqrt{a^2+a+b^2}}$

Therefore;

$\sqrt{a+ib} = > \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$

Answer 1

You made a mistake in your derivation. You should have $(x^2+y^2)^\color{red}2=a^2+b^2$.

This follows from $a=x^2-y^2$ and $b=2xy$ (or from known properties of complex modulus).

Thus, $a=x^2-\dfrac {b^2}{4x^2}$;

solving this quadratic equation in $x^2$ yields $x^2=\dfrac{a+\sqrt{a^2+b^2}}2$ as the correct answer.

Answer 2

One of the approache is use Euler form if Z is a complex numbers such that $ Z\, = \,r.e^{ix}$ then Square root of Z will be $Z_{1}= \, \sqrt{r} . e^{iy} $ Where angle y will be x/2 and (x + 2π)/2.