Q48 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

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

  • Correct answer
    A
    $\tan ^{-1}\left(\frac{1}{3}\right)-\pi$
  • B
    $\tan ^{-1}\left(\frac{3}{4}\right)-\pi$
  • C
    $\pi-\tan ^{-1}\left(\frac{3}{4}\right)$
  • D
    $\tan ^{-1}\left(\frac{1}{3}\right)$

Solution

$\begin{aligned} & \text { } z=\frac{(2-i)(1+i)^3}{(1-i)^2}=\frac{(2-i)\left(1+i^3+3 i(i+1)\right)}{\left(1+i^2-2 i\right)} \\ & =\frac{(2-i)(-2+2 i)}{-2 i}=\frac{(1-i)(2-i)}{i} \\ & =\frac{(1-i)(2 i+1)}{-1}=(-3-i)=z \quad\left(\because \text { lies in } 3^{\text {rd }} \text { quadrant }\right) \\ & \operatorname{Arg}(z)=-\pi+\tan ^{-1}\left(\frac{1}{3}\right)\end{aligned}$

Hence, the correct answer is option: 'A'.

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