Let z,w be complex number such that z bar + i (w bar) = 0 and argzw=​π.Find arg z.

Hi Sagar,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Points} \mathrm{to} \mathrm{remember}: \\ \to \arg \left({\mathrm{z}}_1{\mathrm{z}}_2\right)=\arg \left({\mathrm{z}}_1\right)+\arg \left({\mathrm{z}}_2\right) \\ \to \arg \left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right)=\arg \left({\mathrm{z}}_1\right)-\arg \left({\mathrm{z}}_2\right) \\ \mathrm{Given} \mathrm{that}: \\ \overline{\mathrm{z}}+\mathrm{i}\overline{\mathrm{w}}=0 \\ \Rightarrow \overline{\mathrm{z}}=-\mathrm{i}\overline{\mathrm{w}} \\ \mathrm{Taking} \mathrm{conjugate} \mathrm{on} \mathrm{both} \mathrm{sides} \mathrm{we} \mathrm{get}, \\ \overline{\overline{\mathrm{z}}}=\left(-\overline{\mathrm{i}}\right)\overline{\overline{\mathrm{w}}} \\ \mathrm{We} \mathrm{know} \mathrm{that} \overline{\overline{\mathrm{X}}}=\mathrm{X} \mathrm{and} -\overline{\mathrm{i}}=\mathrm{i} \\ \Rightarrow \mathrm{z}=\mathrm{iw} \\ \Rightarrow \frac{\mathrm{z}}{\mathrm{w}}=\mathrm{i} \\ \Rightarrow \frac{\mathrm{z}}{\mathrm{w}} \mathrm{is} \mathrm{purely} \mathrm{imaginary}. \\ \Rightarrow \arg \left(\frac{\mathrm{z}}{\mathrm{w}}\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \arg \left(\mathrm{z}\right)-\arg \left(\mathrm{w}\right)=\frac{\mathrm{\pi }}{2} ;\mathrm{equation}\left(\mathrm{i}\right) \\ \mathrm{Given} \mathrm{that} \\ \arg \left(\mathrm{zw}\right)=\mathrm{\pi } \\ \Rightarrow \arg \left(\mathrm{z}\right)+\arg \left(\mathrm{w}\right)=\mathrm{\pi } ;\mathrm{equation}\left(\mathrm{ii}\right) \\ \mathrm{equation}\left(\mathrm{i}\right)+\mathrm{equation}\left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}, \\ \arg \left(\mathrm{z}\right)-\arg \left(\mathrm{w}\right)+\arg \left(\mathrm{z}\right)+\arg \left(\mathrm{w}\right)=\frac{\mathrm{\pi }}{2}+\mathrm{\pi } \\ \Rightarrow 2\arg \left(\mathrm{z}\right)=\frac{3\mathrm{\pi }}{2} \\ \mathbf{\Rightarrow }\mathbf{arg}\left(\mathbf{z}\right)\mathbf{=}\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{4}}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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