Answer to Question #316162 in Complex Analysis for Lipika

Question

Find square root of 1+i

Answer 1

$\sqrt{1+i}=a+bi$

$1+i=(a+bi)^2$

$1+i=(a^2-b^2)+2abi$

Equating real and imaginary parts gives us;

$1=a^2-b^2$

1=2ab

This is a simultaneous equation and so we are going to solve by substituting;

$a=\frac{1}{2b}$

$1=(\frac{1}{2b})^2-b^2=\frac{1}{4b^2}-b^2$

$4b^2=1-4b^4$

$4b^4+4b^2-1=0$

$b^2=\frac{-4 \plusmn \sqrt{4^2-4(4)(-1)}}{2(4)}=\frac{-1 \plusmn \sqrt{2}}{2}$

$b^2=\frac{-1-\sqrt{2}}{2}$ has no real roots so we take the positive value

$b=\plusmn \frac{\sqrt{2(\sqrt{2}-1)}}{2}$

$a= \plusmn \frac{2\sqrt{\sqrt{2}-1}+\sqrt{2\sqrt{2}-1}}{2}$

So finally the two possible roots of the complex number 1+i are ;

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