Question

Find the square root of following:
(i) $3+4i$
(ii) $-5+12i$

Answer 1

(i) $3+4i$

$=3+\mathrm{2.2.}i$

$=2^2+\mathrm{2.2.}i+i^2$ [Since $i^2=-1$]

$=(2+i{)}^2$.

Now square root of $(3+4i)$ is $\pm (2+i)$.
(ii) $-5+12i$

$=-5+2.3i.2$

$=(3i{)}^2+2.3i.2+2^2$

$=(3i+2{)}^2$

Now square root of $-5+12i$ is $\pm (3i+2)$.