The square root of 3 – 4i is

Let $\sqrt{3-4i}=x+iy$ $\Rightarrow 3-4i=x^2-y^2+2ixy$

$\Rightarrow x^2-y^2=3, 2xy=-4$      ......(i)

$\Rightarrow {(x^2+y^2)}^2={(x^2-y^2)}^2+4x^2{y}^2$ $={\left(3\right)}^2+{(-4)}^2$= 25 

$\Rightarrow x^2+y^2=5$                  …..(ii)

From equation (i) and (ii) $x^2=4\Rightarrow x=\pm 2$,

$y^2=1\Rightarrow y=\pm 1.$ Hence the square root of  $(3-4i)$ is  $\pm (2+i).$