Question

If ${\left(1+i\sqrt{3}\right)}^9=a+ib$, then $b$ is equal to

• A. $1$
• B. $256$
• C. $0$
• D. $9^3$

Answer 1

The correct option is C

$0$

Explanation for the correct option:

STEP 1: Multiply and divide ${(1+\mathrm{i}\sqrt{3})}^9$ by $2^9$.

Given, ${\left(1+i\sqrt{3}\right)}^9=a+ib$

$\begin{array}{rcl}{\left(1+i\sqrt{3}\right)}^9& =& \frac{2^9{\left(1+i\sqrt{3}\right)}^9}{2^9}\\ & =& 2^9{\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^9\\ & =& 2^9{\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)}^9\end{array}$

STEP 2: Apply De Moivre's theorem.

$\begin{array}{rcl}{2}^9{\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)}^9& =& 2^9\left(\cos \frac{9\pi }{3}+i\sin \frac{9\pi }{3}\right)\\ & =& 2^9\left(\cos 3\pi +i\sin 3\pi \right)\\ & =& 2^9\left(\cos \pi +i\sin \pi \right)\\ & =& 2^9\left(-1+0\right)\\ & =& -2^9\end{array}$

STEP 3: Equating values of ${(1+\mathrm{i}\sqrt{3})}^9$ obtained in step 1 and step 2 to find $a,b$.

$\begin{array}{rcl}{\left(1+i\sqrt{3}\right)}^9& =& a+ib\\ & =& -2^9+0i\end{array}$

$\Rightarrow$ $a=-2^9$ and $b=0$

$\mathbf{\therefore }\mathit{b}\mathbf{=}\mathbf{0}$

Hence option C is correct.