If the equation ${{\left( 1+i\sqrt{3} \right)}^{9}}=a+ib$ holds true, then b is equal to
(a) 1
(b) 256
(c) 0
(d) ${{9}^{3}}$

Hint:Start by converting the complex number $1+i\sqrt{3}$ to its complex form, i.e., to the form $r\left( \cos \theta +i\sin \theta \right)$ where r is the modulus and $\theta $ is the argument. Now further convert it to its exponential form, i.e., $r{{e}^{i\theta }}$ form and substitute in the equation given in the question. Convert the LHS back to polar form and equate the imaginary of LHS with b and give a report the answer.

Complete step-by-step solution -
Let us start by converting the complex number $1+i\sqrt{3}$ to its complex form, i.e., to the form $r\left( \cos \theta +i\sin \theta \right)$ where r is the modulus and $\theta $ is the argument. To convert it to this form, we will divide and multiply the expression by 2. On doing so, we get
$2\left( \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right)$
Now we know that $\cos \dfrac{\pi }{3}=\dfrac{1}{2}\text{ and sin}\dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$ . Using this in our expression, we get
$2\left( \cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3} \right)$
So, for this r=2 and $\theta =\dfrac{\pi }{3}$ . So, if we convert it to exponential form, i.e., $r{{e}^{i\theta }}$ form we get
$2\left( \cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3} \right)=2{{e}^{i\dfrac{\pi }{3}}}$
So, we can say that $1+i\sqrt{3}=2{{e}^{i\dfrac{\pi }{3}}}$ .So, if we substitute this in the equation given in the question, we get
${{\left( 1+i\sqrt{3} \right)}^{9}}=a+ib$
$\Rightarrow {{\left( 2{{e}^{i\dfrac{\pi }{3}}} \right)}^{9}}=a+ib$
$\Rightarrow {{2}^{9}}{{e}^{i\dfrac{\pi }{3}\times 9}}=a+ib$
$\Rightarrow {{2}^{9}}{{e}^{i\left( 3\pi \right)}}=a+ib$
For the complex number ${{2}^{9}}{{e}^{i\left( 3\pi \right)}}$ , $r={{2}^{9}}$ and $\theta =3\pi $ . So, if we convert it to polar form, we get
${{2}^{9}}\left( \cos 3\pi +i\sin 3\pi \right)=a+ib$
Now we know that the value of sine for odd multiple of $\pi $ is 0 and the value of cosine for odd multiple of $\pi $ is -1.
${{2}^{9}}\left( -1+i\times 0 \right)=a+ib$
$\Rightarrow -{{2}^{9}}=a+ib$
As the imaginary part of the LHS is zero, so the value of b is equal to 0.
Hence, the answer to the above question is option (c).

Note: Remember that the modulus r of the complex number x+iy is given by $\sqrt{{{x}^{2}}+{{y}^{2}}}$ and $\sin \theta =\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\text{ and cos}\theta \text{=}\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$ . Also, the signs of $\sin \theta \text{ and cos}\theta $ decides the quadrant in which the argument lies. The argument and modulus for the polar form and exponential forms are the same. If you want you can remember the formula that $\left( \operatorname{cosA}+isinA \right)\left( \cos B+i\sin B \right)\left( \operatorname{cosC}+i\operatorname{sinC} \right)=\cos \left( A+B+C \right)+i\sin \left( A+B+C \right)$ , so for the above question you can directly ${{\left( 1+i\sqrt{3} \right)}^{9}}={{2}^{9}}{{\left( \cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3} \right)}^{9}}={{2}^{9}}\left( \cos \dfrac{9\times \pi }{3}+i\sin \dfrac{9\times \pi }{3} \right)$ , removing the step of converting it to exponential form.