Question

$\sqrt{-35+12i}-\sqrt{-35-12i}=$

• A. 6i
• B. 12i
• C. 4i
• D. 8i

Answer 1

The correct option is B 12i

Let $(a+ib{)}^2=-35+12i$

Hence,

$a^2-b^2=-35$

Also, $a^2+b^2=\sqrt{{35}^2+{12}^2}$

$a^2+b^2=37$

Hence, $a^2=1$

$b^2=36$

Also, $ab=6$

Hence, $a=\pm 1,b=\pm 6$ . Note : $a,b$ will be of same signs.

Similarly,

Let $(c+id{)}^2=-35-12i$

On solving we will get, $c=\pm 1,d=\pm 6$ Note: $c,d$ will be of opposite signs.

Hence, the required answer $=(1+6i)-(1-6i)=12i$ or $=(1+6i)-(-1+6i)=2$ or $(-1-6i)-(1-6i)=-2$ or $=(-1-6i)-(-1+6i)=-12i$

Hence, option B is correct.