Question

If |z| = 4 and $\arg \left(\mathrm{z}\right)=\frac{5\mathrm{\pi }}{6},$ then z = ____________.

Answer 1

If |z| = 4 and $\arg \mathrm{z}=\frac{5\mathrm{\pi }}{6}=\frac{\mathrm{\pi }-\mathrm{\pi }}{6}$
$$\begin{gathered} \therefore z=\left|z\right|\left[\cos \left(\frac{5\mathrm{\pi }}{6}\right)+i\sin \left(\frac{5\mathrm{\pi }}{6}\right)\right] \\ =4\left[\cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)+i\sin \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)\right] \\ \left[\mathrm{Since}\cos \left(\mathrm{\pi }-\theta \right)=-\cos \theta \mathrm{and}\sin \left(\mathrm{\pi }-\theta \right)=\sin \theta \right] \\ \therefore z=4\left[-\cos \frac{\mathrm{\pi }}{6}+i\sin \frac{\mathrm{\pi }}{6}\right] \\ \mathrm{Since}\cos \frac{\mathrm{\pi }}{6}=\frac{\sqrt{3}}{2} \\ \sin \frac{\mathrm{\pi }}{6}=\frac{1}{2} \\ \Rightarrow z=4\left[-\frac{\sqrt{3}}{2}+\frac{i}{2}\right]=2\left(-\sqrt{3}+i\right) \\ \mathrm{Hence},z=-2\sqrt{3}+2i \end{gathered}$$