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TS EAMCET · Maths · Complex Number

If \(z=\frac{(2-i)(1+i)^3}{(1-i)^2}\), then \(\operatorname{Arg}(z)=\)

  1. A \(\tan ^{-1}\left(\frac{1}{3}\right)-\pi\)
  2. B \(\tan ^{-1}\left(\frac{3}{4}\right)-\pi\)
  3. C \(\pi-\tan ^{-1}\left(\frac{3}{4}\right)\)
  4. D \(\tan ^{-1}\left(\frac{1}{3}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\tan ^{-1}\left(\frac{1}{3}\right)-\pi\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \text { } z=\frac{(2-i)(1+i)^3}{(1-i)^2}=\frac{(2-i)\left(1+i^3+3 i(i+1)\right)}{\left(1+i^2-2 i\right)} \\ & =\frac{(2-i)(-2+2 i)}{-2 i}=\frac{(1-i)(2-i)}{i} \\ & =\frac{(1-i)(2 i+1)}{-1}=(-3-i)=z \quad\left(\because \text { lies in } 3^{\text {rd }} \text {…