Q100 / 180 · COMEDK 2024 (Memory Based) · Mathematics
If $(1+i)^{200}=2^{98}(a+i b)$ then the value of $a+b$ is
- A2
- Correct answerB4
- C6
- D8
Solution
We have, $(1+i)^{200}$
Multiply and divide by $\sqrt{2}$
$\begin{aligned}<br/>& {\left[\sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\right]^{200}=\sqrt{2}^{200}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{200}} \\<br/>& =2^{100}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{200}<br/>\end{aligned}$
$\begin{aligned}<br/>& \Rightarrow(1+i)^{200}=2^{100}=2^{95}(a+i b) \\<br/>& 2^2=a+i b \\<br/>& 4+i 0=a+i b<br/>\end{aligned}$
On comparing $a=4, b=0$
$\therefore \quad a+b=4$
Multiply and divide by $\sqrt{2}$
$\begin{aligned}<br/>& {\left[\sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\right]^{200}=\sqrt{2}^{200}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{200}} \\<br/>& =2^{100}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{200}<br/>\end{aligned}$
$\begin{aligned}<br/>& \Rightarrow(1+i)^{200}=2^{100}=2^{95}(a+i b) \\<br/>& 2^2=a+i b \\<br/>& 4+i 0=a+i b<br/>\end{aligned}$
On comparing $a=4, b=0$
$\therefore \quad a+b=4$
Hence, the correct answer is option: 'B'.