Q48 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics

If $z$ and $w$ are complex numbers such that $\bar{z}-i \bar{w}=0$ and $\operatorname{Arg}(z w)=\frac{3 \pi}{4}$, then $\operatorname{Arg} z=$

  • A
    $\frac{\pi}{16}$
  • Correct answer
    B
    $\frac{\pi}{8}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{3 \pi}{4}$

Solution

We have, $$ \begin{aligned} \Rightarrow & i \bar{\omega} & =\bar{z} \Rightarrow \omega=\frac{1}{i} \bar{z} \\ \Rightarrow & \omega & =-\frac{1}{i} z \Rightarrow \omega=i z \end{aligned} $$ Now, we have $$ \begin{aligned} & \arg (z \omega)=\frac{3 \pi}{4} \\ & \Rightarrow \quad \arg (z(i z))=\frac{3 \pi}{4} \\ & \Rightarrow \quad \arg \left(i z^2\right)=\frac{3 \pi}{4} \\ & \Rightarrow \quad \arg (i)+\arg \left(z^2\right)=\frac{3 \pi}{4} \\ & {\left[\because \arg \left(z_1 z_2\right)=\arg \left(z_1\right)+\arg \left(z_2\right)\right]} \\ & \Rightarrow \quad \arg (i)+2 \arg (z)=\frac{3 \pi}{4} \quad\left[\because \arg \left(z^n\right)=n \arg (z)\right] \\ & \Rightarrow \quad \frac{\pi}{2}+2 \arg (z)=\frac{3 \pi}{4} \\ & \Rightarrow \quad \arg (z)=\frac{\pi}{8} \\ & \end{aligned} $$

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

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