Search any question & find its solution
Question:
Answered & Verified by Expert
If $(\sqrt{3}+i)^{100}=2^{99}(a+i b)$, then $a^{2}+b^{2}$ is equal to
Options:
Solution:
2794 Upvotes
Verified Answer
The correct answer is:
4
$(\sqrt{3}+i)^{100}=2^{99}(a+i b) \Rightarrow\left(-2 i(0)^{100}=2^{99}(a+i b)\right.$
$$
\begin{aligned}
&{\left[\because \omega=-\frac{1+\sqrt{3} i}{2}, i \omega=-\frac{i-\sqrt{3}}{2}\right.} \\
&\left.\Rightarrow \quad-2 i \omega=i+\sqrt{3} \text { and } \omega^{100}=\omega_{\bullet}^{99} \omega=1 \cdot \omega=\omega\right] \\
&\Rightarrow 2\left\{\frac{-1-\sqrt{3 i}}{2}\right\}=a+i b \Rightarrow a=-1 \text { and } b=\sqrt{3} \\
&\therefore a^{2}+b^{2}=(-1)^{2}+(\sqrt{3})^{2}=4
\end{aligned}
$$
$$
\begin{aligned}
&{\left[\because \omega=-\frac{1+\sqrt{3} i}{2}, i \omega=-\frac{i-\sqrt{3}}{2}\right.} \\
&\left.\Rightarrow \quad-2 i \omega=i+\sqrt{3} \text { and } \omega^{100}=\omega_{\bullet}^{99} \omega=1 \cdot \omega=\omega\right] \\
&\Rightarrow 2\left\{\frac{-1-\sqrt{3 i}}{2}\right\}=a+i b \Rightarrow a=-1 \text { and } b=\sqrt{3} \\
&\therefore a^{2}+b^{2}=(-1)^{2}+(\sqrt{3})^{2}=4
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.