Exercise 13.4

If $\mathbf{z}_{1}$ and $\mathbf{z}_{2}$ are two complex number such that $\left|\mathbf{z}_{1}\right|=\left|\mathbf{z}_{2}\right|$ and $\arg \left(\mathbf{z}_{1}\right)+\arg \left(\mathbf{z}_{2}\right)=\pi$, then show that $\mathbf{z}_{1}=-\overline{\mathbf{z}_{2}}$

Solution: Given that $\left|z_{1}\right|=\left|z_{2}\right|$ and $\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=\pi$ Let's assume $\arg \left(z_{1}\right)=\theta$ $\arg...

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