(v) Given ∣z∣=103|z|=\dfrac{10}{3}∣z∣=310 and arg(z)=−17π12\arg(z)=-\dfrac{17\pi}{12}arg(z)=−1217π. z=103(cos(−17π12)+isin(−17π12))z=\frac{10}{3}\left(\cos\left(-\frac{17\pi}{12}\right)+i\sin\left(-\frac{17\pi}{12}\right)\right)z=310(cos(−1217π)+isin(−1217π)) −17π12=−π−5π12-\frac{17\pi}{12}=-\pi-\frac{5\pi}{12}−1217π=−π−125π cos(−17π12)=cos(π+5π12)=−cos5π12=2−64\cos\left(-\frac{17\pi}{12}\right)=\cos\left(\pi+\frac{5\pi}{12}\right)=-\cos\frac{5\pi}{12}=\frac{\sqrt{2}-\sqrt{6}}{4}cos(−1217π)=cos(π+125π)=−cos125π=42−6 sin(−17π12)=−sin(π+5π12)=sin5π12=6+24\sin\left(-\frac{17\pi}{12}\right)=-\sin\left(\pi+\frac{5\pi}{12}\right)=\sin\frac{5\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4}sin(−1217π)=−sin(π+125π)=sin125π=46+2 z=5(2−6)6+5(6+2)6i\boxed{z=\frac{5(\sqrt{2}-\sqrt{6})}{6}+\frac{5(\sqrt{6}+\sqrt{2})}{6}i}z=65(2−6)+65(6+2)i