Question

If ${(\sqrt{3}+i)}^{100}=2^{99}(a+ib)$, then $a^2+b^{2}is$

• A. $\sqrt{2}$
• B. $4$
• C. $\sqrt{3}$
• D. None of these

Answer 1

The correct option is B

$4$

Explanation for the correct option:

Step 1 : Solve using the conjugate of complex number

Given ${(\sqrt{3}+i)}^{100}=2^{99}(a+ib).......\left(i\right)$,

Therefore, its conjugate complex number ${(\sqrt{3}-i)}^{100}=2^{99}(a-ib)......\left(ii\right)$

Multiply both the above equation as follows

$\begin{array}{c}{(\sqrt{3}+i)}^{100}\cdot {(\sqrt{3}-i)}^{100}=2^{99}(a+ib)\cdot 2^{99}(a-ib)\\ {\Rightarrow \left[(\sqrt{3}+i)(\sqrt{3}-i)\right]}^{100}=2^{99+99}(a+ib)\cdot (a-ib)\\ {\Rightarrow (3+1)}^{100}=2^{198}(a^2+b^2)\\ \Rightarrow 4^{100}=2^{198}(a^2+b^2)\\ \Rightarrow (a^2+b^2)=\frac{2^{200}}{2^{198}}\\ \Rightarrow a^2+b^2=2^2\\ \Rightarrow a^2+b^2=4\end{array}$

In another way,

Step 2 : Solve using the known value of $\omega$

Since $\omega =\frac{-1+i\sqrt{3}}{2}$, then

$\begin{array}{l}\Rightarrow \omega =\frac{i^2+i\sqrt{3}}{2}\\ \Rightarrow \frac{2\omega }{i}=(\sqrt{3}+i)\\ \Rightarrow (\sqrt{3}+i)=-2\omega i\end{array}$

Take the power of $100$ on both the sides, then

$\begin{array}{c}{(\sqrt{3}+i)}^{100}={(-2\omega i)}^{100}\\ =2^{100}{\omega }^{100}{i}^{100}\\ =2^{100}\omega \cdot 1\\ =2^{100}\left(\frac{-1+i\sqrt{3}}{2}\right)\\ =2^{99}(-1+i\sqrt{3})\end{array}$

Comparing with the given equation, obtain $a=-1$ and $b=\sqrt{3}$, then

$a^2+b^2=4$

Hence, the correct option is (B).