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KCET · Maths · Complex Number

If \( \left(\frac{1-i}{1+i}\right)^{96}=a+i b \) then \( (a, b) \) is

  1. A \( (1,1) \)
  2. B \( (1,0) \)
  3. C \( (0,1) \)
  4. D \( (0,-1) \)
Verified Solution

Answer & Solution

Correct Answer

(B) \( (1,0) \)

Step-by-step Solution

Detailed explanation

Given that,
\( \left(\frac{1-i}{1+i}\right)^{96}=a+i b \)
Now, \( \frac{1-i}{1+i}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}=-i \)
So, \( i^{96}=a+i b \)
\( \Rightarrow\left(i^{4}\right)^{24}=a+i b \)
\( \Rightarrow 1+i(0)=a+i b \)
Therefore, \( (a, b) \) is \( (1,0) \).