Square root of a complex number

SQUARE ROOT OF A COMPLEX NUMBER

The solution of the equation $$z^2=a+ib$$ is given in two forms.

In terms of $a,b$,
$$z= \pm \left( \sqrt{{ \sqrt{a^2+b^2}+a \over  2}} + i {b\over |b|}\sqrt{{ \sqrt{a^2+b^2}-a \over  2}}\right)$$

In terms of $w=a+ib$,
$$z= \pm \left(\sqrt{{ |w|+\Re(w) \over  2}} + i{\Im(w)\over |\Im(w)|} \sqrt{{ |w|-\Re(w) \over  2}}\right)$$

A crazy person could just check this by squaring it! you should get $w=a+ib$ back.

=========================== WORKING BELOW

Suppose we wish to find the square root of a complex number $w=a+ib$.

Some notation we use later regarding real and imaginary parts of the complex number.

$\Re(w)=a$
$\Im(w)=b$
${x\over |x|}$   gives the sign of $x$, $+1$ or $-1$.

Let $z=x+iy$, and of course keep in mind that we use $i^2=-1$.


$$\begin{array}{ll} z^2 &=a+ib\\ (x+iy)^2 &= a+ib\\x^2-y^2+i2xy &= a+ib  \end{array}$$

Therefore equating real and imaginary parts we have two equations:

$x^2-y^2=a$ and $2xy=b$. 

$x^2-\displaystyle{b^2\over 4x^2}=a$

$4 x^4-b^2=4 x^2 a$

$4x^4-4a x^2 -b^2 = 0$

$x^2 = {4a \pm \sqrt{16a^2-4\times 4(-b^2)} \over 8}$

$x^2 = {4a \pm \sqrt{16a^2+16b^2} \over 8}$
$x^2 = {a \pm \sqrt{a^2+b^2} \over 2}$

$x^2 = {a + \sqrt{a^2+b^2} \over 2}$
 since $x^2\ge 0$ needs to hold.

$\therefore x = \pm \sqrt{{a + \sqrt{a^2+b^2} \over 2}}$

$\therefore y= \pm {b\over 2} \sqrt{{2 \over a+\sqrt{a^2+b^2} }}$

rationalise the denominator and bring the '$b/2$' inside the square root symbol by keeping in mind that $b=sign(b)\times |b|$.

$\therefore y= \pm sign(b) \sqrt{{b^2\over 4}\,{2 \over a+\sqrt{a^2+b^2} }  \times { a-\sqrt{a^2+b^2} \over a-\sqrt{a^2+b^2}}}$

Finally in terms of $a$ and $b$ we get

$$z= \pm \sqrt{{ \sqrt{a^2+b^2}+a \over  2}} \pm i {b\over |b|} \sqrt{{ \sqrt{a^2+b^2}-a \over  2}}$$

or in terms of $w$

$$z= \pm \sqrt{{ |w|+\Re(w) \over  2}} \pm i {\Im(w)\over |\Im(w)|}\sqrt{{ |w|-\Re(w) \over  2}}$$

The $\pm$ means you either take both as $+$ or both as $-$. :-)

Or with a big bracket but looks more cumbersome ..

So in terms of $a,b$,
$$z= \pm \left( \sqrt{{ \sqrt{a^2+b^2}+a \over  2}} + i {b\over |b|} \sqrt{{ \sqrt{a^2+b^2}-a \over  2}}\right)$$

Or in terms of $w$,
$$z= \pm \left(\sqrt{{ |w|+\Re(w) \over  2}} + i {\Im(w)\over |\Im(w)|}\sqrt{{ |w|-\Re(w) \over  2}}\right)$$

A crazy person could just check this by squaring it! you should get $w=a+ib$ back.

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Postscript Note:
Apparently it can also be shown that

$$\sqrt{z}=\sqrt{|z|}\,{ z+|z|\over \left| z + |z| \right| }$$

where $\sqrt{z}$ means the principal square root, which is where the real part is positive.

You can prove this formula by taking $z=x+iy$ and brute force it for example :-), enjoy!!