Question

The square root of $3-4i$ is

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

$\pm (2-i)$

Let $\sqrt{3-4i}=x+iy$

$\Rightarrow 3-4i=x^2-y^2+2ixy$

Lets equate the real parts and the imaginary parts

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

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

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

From equation (i) and (ii)

$x^2+y^2+x^2-y^2=5+3$

$\Rightarrow 2x^2=8$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$---(1)

$x^2+y^2-x^2+y^2=5-3$

$\Rightarrow 2y^2=2$

$\Rightarrow y^2=1$

$\Rightarrow y=\pm 1$

Therefore, $\sqrt{3-4i}=\pm (2+i)$