1708$Is It $\mathbf{\arg(z\omega)=\pi}
Let z and w be two complex numbers such that \bar{z} +i\bar{w} =0 and arg zw=∩ , then arg z is equal to ?
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3 Answers
1708$\textbf{Here $\mathbf{\bar z+i\bar \omega=0\Leftrightarrow \bar z=-i\bar \omega}$}\\\\ \textbf{taking Conjugate on both side,We Get}\\\\ $\mathbf{z=i\omega...........................(1)}$\\\\ $\mathbf{\arg(z\omega)=\pi\Leftrightarrow \arg(z)+\arg(\omega)=\pi.....................(2)}$\\\\ \textbf{Put value of $\mathbf{z}$ from equ....(1), We Get}\\\\ $\mathbf{\arg(i\omega)+\arg(\omega)=\pi\Leftrightarrow \arg(i)+\arg(\omega)+\arg(\omega)=\pi}$\\\\ $\mathbf{\frac{\pi}{2}+2.\arg(\omega)=\pi\Leftrightarrow \arg(\omega)=\frac{\pi}{4}}$\\\\ \textbf{So from eq...(2), We Get $\underline{\underline{\mathbf{\arg(z)=\frac{3\pi}{4}}}}$}