Q.Find the square roots of $3+4i$.

Set $\sqrt{3+4i}=x+yi$; use $x^{2}-y^{2}=3$, $2xy=4$ and $x^{2}+y^{2}=\sqrt{3^{2}+4^{2}}=5$.

Let $\sqrt{3+4i}=x+yi$ with $x,y$ real. Squaring:

Equate real and imaginary parts:

From moduli, $x^{2}+y^{2}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5.$

Add and subtract with $x^{2}-y^{2}=3$:

Since $2xy=4>0$, $x$ and $y$ have the same sign, giving the matched pair …

Reporting only $2+i$ and dropping $-(2+i)$ — a non-zero complex number has TWO square roots. Also, ignoring $2xy=4>0$ and pairing opposi …