Question

Find the square root of $8-6i$. ( $i^2=-1$ )

• A. $3-i$
• B. $3+i$
• C. $3-4i$
• D. $4-3i$
• E. $4+3i$

Answer 1

The correct option is A $3-i$

Let the square root of $8-6i$ be $a+ib$

$\therefore 8-6i=(a+ib{)}^2$

$\therefore 8-6i=a^2-b^2+i\left(2ab\right)$

Comparing the corresponding terms, we get

$8=a^2-b^2$

$-6=2ab$

$\Rightarrow ab=-3,a^2-b^2=8$

$\Rightarrow a^2-\frac{9}{a^2}=8$

$\Rightarrow a^4-8a^2-9=0$

$\therefore a^4-9a^2+a^2-9=0$

$\therefore (a^2+1)(a^2-9)=0$

Since $a^2$ is always positive, $a^2+1\ne 0$

$\therefore a^2-9=0$

$\Rightarrow a=3,b=-1$ or $a=-3,b=1$

The square root is thus $\pm (3-i)$