Let $z$ and $w$ be two complex numbers such that $\bar{z}+i \bar{w}=0$ and $\operatorname{Arg}(z w)=\pi$. Then, $\operatorname{Arg} z=$

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Aicharya · Accepted Answer

Given, $\bar{z}+i \bar{w}=0$ $\Rightarrow \quad \bar{z}-i \bar{w}=0 \quad[\because \bar{i}=-i]$ $\Rightarrow \quad \bar{z}-\overline{i w}=0 \quad\left[\because \overline{z_1 \cdot z_2}=\bar{z}_1 \cdot \bar{z}_2ight]$ $\Rightarrow \quad \overline{z-i w}=0 \quad\left[\because \bar{z}_1-\bar{z}_2=\bar{z}_1-\bar{z}_2ight]$ $\Rightarrow \quad z-i w=0$ $z=i w$ According to question, $\arg (z w)=\pi$ $\begin{array}{lll}\Rightarrow & \arg z+\arg w=\pi & \ \Rightarrow & \arg z+\arg \left(\frac{z}{i}ight)=\pi \quad[\because z=i w]\end{array}$ $\Rightarrow \quad \arg z+\arg z-\arg (i)=\pi$ $2 \arg (z)-\frac{\pi}{2}=\pi$ $\Rightarrow \quad \arg (z)=\frac{3 \pi}{4}$

Let $z$ and $w$ be two complex numbers such that $\bar{z}+i \bar{w}=0$ and $\operatorname{Arg}(z w)=\pi$.…

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