Question

If $x=1+2i$ and $A=x^3+7x^2-x+26$, then one of the value of $\sqrt{A}$ equals

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

Answer 1

Answer: (D)

$3+4i$

$x=1+2i\Rightarrow x^2-2x+5=0$
now $A=x^3+7x^2-x+26$
$=x(x^2-2x+5)+9(x^2-2x+5)+12x-19=12(1+2i)-19=-7+24i$
$\therefore \sqrt{A}=\sqrt{-7+24i}$
$=\pm [\sqrt{\frac{\left|A\right|+Re\left(A\right)}{2}}+i\sqrt{\frac{\left|A\right|-Re\left(A\right)}{2}}]=\pm (3+4i)$