\displaystyle \textbf{Question 1: }\text{Find the modulus and argument of the following complex numbers and hence}
\displaystyle \text{express each one of them in polar form:}
\displaystyle \text{(i) }1+i\qquad\text{(ii) }\sqrt{3}+i\qquad\text{(iii) }1-i\qquad\text{(iv) }\frac{1-i}{1+i}
\displaystyle \text{(v) }\frac{1}{1+i}\qquad\text{(vi) }\frac{1+2i}{1-3i}\qquad\text{(vii) }\sin120^\circ-i\cos120^\circ
\displaystyle \text{(viii) }\frac{-16}{1+i\sqrt{3}}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }z=1+i
\displaystyle |z|=\sqrt{1^2+1^2}=\sqrt{2}
\displaystyle \text{Let }\tan\alpha=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|
\displaystyle \Rightarrow \tan\alpha=\left|\frac{1}{1}\right|=1
\displaystyle \Rightarrow \alpha=\frac{\pi}{4}
\displaystyle \text{Since the point }(1,1)\text{ lies in the first quadrant,}
\displaystyle \arg(z)=\theta=\frac{\pi}{4}
\displaystyle \therefore \text{The polar form of }1+i\text{ is}
\displaystyle z=r(\cos\theta+i\sin\theta)
\displaystyle =\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right).
\displaystyle \\

\displaystyle \text{(ii) }z=\sqrt{3}+i
\displaystyle |z|=\sqrt{(\sqrt{3})^2+1^2}=2
\displaystyle \text{Let }\tan\alpha=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|
\displaystyle \Rightarrow \tan\alpha=\left|\frac{1}{\sqrt{3}}\right|=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \alpha=\frac{\pi}{6}
\displaystyle \text{Since the point }(\sqrt{3},1)\text{ lies in the first quadrant,}
\displaystyle \arg(z)=\theta=\frac{\pi}{6}
\displaystyle \therefore \text{The polar form of }\sqrt{3}+i\text{ is}
\displaystyle z=r(\cos\theta+i\sin\theta)
\displaystyle =2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right).
\displaystyle \\

\displaystyle \text{(iii) }z=1-i
\displaystyle |z|=\sqrt{1^2+(-1)^2}=\sqrt{2}
\displaystyle \text{Let }\tan\alpha=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|
\displaystyle \Rightarrow \tan\alpha=\left|\frac{-1}{1}\right|=1
\displaystyle \Rightarrow \alpha=\frac{\pi}{4}
\displaystyle \text{Since the point }(1,-1)\text{ lies in the fourth quadrant,}
\displaystyle \arg(z)=\theta=-\alpha=-\frac{\pi}{4}
\displaystyle \therefore \text{The polar form of }1-i\text{ is}
\displaystyle z=r(\cos\theta+i\sin\theta)
\displaystyle =\sqrt{2}\left(\cos\frac{-\pi}{4}+i\sin\frac{-\pi}{4}\right)
\displaystyle =\sqrt{2}\left(\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}\right).
\displaystyle \\

\displaystyle \text{(iv) }z=\frac{1-i}{1+i}
\displaystyle =\frac{1-i}{1+i}\times\frac{1-i}{1-i}
\displaystyle =\frac{(1-i)^2}{1-i^2}
\displaystyle =\frac{1-2i+i^2}{2}
\displaystyle =\frac{-2i}{2}=-i
\displaystyle \therefore z=-i
\displaystyle |z|=\sqrt{0^2+(-1)^2}=1
\displaystyle \text{Since the point }(0,-1)\text{ lies on the negative imaginary axis,}
\displaystyle \arg(z)=\theta=-\frac{\pi}{2}
\displaystyle \therefore \text{The polar form of }\frac{1-i}{1+i}\text{ is}
\displaystyle z=r(\cos\theta+i\sin\theta)
\displaystyle =\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)
\displaystyle =\cos\frac{\pi}{2}-i\sin\frac{\pi}{2}.
\displaystyle \\

\displaystyle \text{(v) }z=\frac{1}{1+i}
\displaystyle =\frac{1}{1+i}\times\frac{1-i}{1-i}
\displaystyle =\frac{1-i}{1-i^2}
\displaystyle =\frac{1-i}{2}=\frac{1}{2}-\frac{1}{2}i
\displaystyle \therefore z=\frac{1}{2}-\frac{1}{2}i
\displaystyle |z|=\sqrt{\left(\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)^2}
\displaystyle =\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}
\displaystyle \text{Let }\tan\alpha=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|
\displaystyle \Rightarrow \tan\alpha=\left|\frac{-\frac{1}{2}}{\frac{1}{2}}\right|=1
\displaystyle \Rightarrow \alpha=\frac{\pi}{4}
\displaystyle \text{Since the point }\left(\frac{1}{2},-\frac{1}{2}\right)\text{ lies in the fourth quadrant,}
\displaystyle \arg(z)=\theta=-\alpha=-\frac{\pi}{4}
\displaystyle \therefore \text{The polar form of }\frac{1}{1+i}\text{ is}
\displaystyle z=r(\cos\theta+i\sin\theta)
\displaystyle =\frac{1}{\sqrt{2}}\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)
\displaystyle =\frac{1}{\sqrt{2}}\left(\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}\right).
\displaystyle \\

\displaystyle \text{(vi) }z=\frac{1+2i}{1-3i}
\displaystyle =\frac{1+2i}{1-3i}\times\frac{1+3i}{1+3i}
\displaystyle =\frac{(1+2i)(1+3i)}{1-(3i)^2}
\displaystyle =\frac{1+3i+2i+6i^2}{10}
\displaystyle =\frac{-5+5i}{10}=-\frac{1}{2}+\frac{1}{2}i
\displaystyle \therefore z=-\frac{1}{2}+\frac{1}{2}i
\displaystyle |z|=\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}
\displaystyle =\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}
\displaystyle \text{Let }\tan\alpha=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|
\displaystyle \Rightarrow \tan\alpha=\left|\frac{\frac{1}{2}}{-\frac{1}{2}}\right|=1
\displaystyle \Rightarrow \alpha=\frac{\pi}{4}
\displaystyle \text{Since the point }\left(-\frac{1}{2},\frac{1}{2}\right)\text{ lies in the second quadrant,}
\displaystyle \arg(z)=\theta=\pi-\alpha
\displaystyle =\pi-\frac{\pi}{4}=\frac{3\pi}{4}
\displaystyle \therefore \text{The polar form of }\frac{1+2i}{1-3i}\text{ is}
\displaystyle z=r(\cos\theta+i\sin\theta)
\displaystyle =\frac{1}{\sqrt{2}}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right).
\displaystyle \\

\displaystyle \text{(vii) }z=\sin120^\circ-i\cos120^\circ=\frac{\sqrt{3}}{2}+\frac{1}{2}i
\displaystyle |z|=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{1}=1
\displaystyle \text{Let }\tan\alpha=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|
\displaystyle \Rightarrow \tan\alpha=\left|\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right|=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \alpha=\frac{\pi}{6}
\displaystyle \text{Since the point }\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\text{ lies in the first quadrant,}
\displaystyle \arg(z)=\theta=\frac{\pi}{6}
\displaystyle \therefore \text{The polar form of }\sin120^\circ-i\cos120^\circ\text{ is}
\displaystyle z=r(\cos\theta+i\sin\theta)
\displaystyle =\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}.
\displaystyle \\

\displaystyle \text{(viii) }z=\frac{-16}{1+i\sqrt{3}}
\displaystyle =\frac{-16}{1+i\sqrt{3}}\times\frac{1-i\sqrt{3}}{1-i\sqrt{3}}
\displaystyle =\frac{-16+16\sqrt{3}i}{4}
\displaystyle =-4+4\sqrt{3}i
\displaystyle \therefore z=-4+4\sqrt{3}i
\displaystyle |z|=\sqrt{(-4)^2+(4\sqrt{3})^2}=\sqrt{16+48}=\sqrt{64}=8
\displaystyle \text{Let }\tan\alpha=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|
\displaystyle \Rightarrow \tan\alpha=\left|\frac{4\sqrt{3}}{-4}\right|=\sqrt{3}
\displaystyle \Rightarrow \alpha=\frac{\pi}{3}
\displaystyle \text{Since the point }(-4,\,4\sqrt{3})\text{ lies in the second quadrant,}
\displaystyle \arg(z)=\theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}
\displaystyle \therefore \text{The polar form of }\frac{-16}{1+i\sqrt{3}}\text{ is}
\displaystyle z=r(\cos\theta+i\sin\theta)
\displaystyle =8\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right).
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write }(i^{25})^3\text{ in polar form.}
\displaystyle \text{Answer:}
\displaystyle z=(i^{25})^3=i^{75}=i^{4\times18+3}=i^3=-i
\displaystyle \therefore z=-i
\displaystyle |z|=\sqrt{0^2+(-1)^2}=1
\displaystyle \text{Since the point }(0,-1)\text{ lies on the negative imaginary axis,}
\displaystyle \arg(z)=\theta=-\frac{\pi}{2}
\displaystyle \therefore \text{The polar form of }(i^{25})^3\text{ is}
\displaystyle z=r(\cos\theta+i\sin\theta)
\displaystyle =\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)
\displaystyle =\cos\frac{\pi}{2}-i\sin\frac{\pi}{2}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Express the following complex numbers in the form }r(\cos\theta+i\sin\theta):
\displaystyle \text{(i) }1+i\tan\alpha\qquad\text{(ii) }\tan\alpha-i\qquad\text{(iii) }1-\sin\alpha+i\cos\alpha
\displaystyle \text{(iv) }\frac{1-i}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }z=1+i\tan\alpha.
\displaystyle \tan\alpha\text{ is a periodic function with period }\pi.
\displaystyle \text{Hence, we may take }\alpha\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right].
\displaystyle \underline{\text{Case I: }}\alpha\in\left[0,\frac{\pi}{2}\right)
\displaystyle z=1+i\tan\alpha
\displaystyle |z|=\sqrt{1+\tan^2\alpha}
\displaystyle =\sqrt{\sec^2\alpha}=|\sec\alpha|=\sec\alpha
\displaystyle \text{Let }\beta\text{ be the reference angle such that}
\displaystyle \tan\beta=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|.
\displaystyle \tan\beta=|\tan\alpha|=\tan\alpha
\displaystyle \Rightarrow \beta=\alpha
\displaystyle \text{For }0<\alpha<\frac{\pi}{2},\text{ we have }\text{Re}(z)>0\text{ and }\text{Im}(z)>0.
\displaystyle \text{Therefore, }z\text{ lies in the first quadrant and }\arg(z)=\alpha.
\displaystyle \text{When }\alpha=0,\ z=1\text{ and }\arg(z)=0.
\displaystyle \therefore z=\sec\alpha(\cos\alpha+i\sin\alpha).
\displaystyle \underline{\text{Case II: }}\alpha\in\left(\frac{\pi}{2},\pi\right]
\displaystyle z=1+i\tan\alpha
\displaystyle |z|=\sqrt{1+\tan^2\alpha}
\displaystyle =\sqrt{\sec^2\alpha}=|\sec\alpha|=-\sec\alpha
\displaystyle \text{Let }\beta\text{ be the reference angle such that}
\displaystyle \tan\beta=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|.
\displaystyle \tan\beta=|\tan\alpha|=-\tan\alpha
\displaystyle =\tan(\pi-\alpha)
\displaystyle \Rightarrow \beta=\pi-\alpha
\displaystyle \text{For }\frac{\pi}{2}<\alpha<\pi,\text{ we have }\text{Re}(z)>0\text{ and }\text{Im}(z)<0.
\displaystyle \text{Therefore, }z\text{ lies in the fourth quadrant.}
\displaystyle \arg(z)=-\beta=-(\pi-\alpha)=\alpha-\pi
\displaystyle \text{When }\alpha=\pi,\ z=1\text{ and }\arg(z)=0=\alpha-\pi.
\displaystyle \therefore z=-\sec\alpha\left(\cos(\alpha-\pi)+i\sin(\alpha-\pi)\right).
\displaystyle \\

\displaystyle \text{(ii) Let }z=\tan\alpha-i.
\displaystyle \tan\alpha\text{ is a periodic function with period }\pi.
\displaystyle \text{Hence, we may take }\alpha\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right].
\displaystyle \underline{\text{Case I: }}\alpha\in\left[0,\frac{\pi}{2}\right)
\displaystyle z=\tan\alpha-i
\displaystyle |z|=\sqrt{\tan^2\alpha+1}
\displaystyle =\sqrt{\sec^2\alpha}=|\sec\alpha|=\sec\alpha
\displaystyle \text{Let }\beta\text{ be the reference angle such that}
\displaystyle \tan\beta=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|.
\displaystyle \tan\beta=\left|\frac{-1}{\tan\alpha}\right|=|\cot\alpha|=\cot\alpha
\displaystyle =\tan\left(\frac{\pi}{2}-\alpha\right)
\displaystyle \Rightarrow \beta=\frac{\pi}{2}-\alpha
\displaystyle \text{For }0<\alpha<\frac{\pi}{2},\text{ we have }\text{Re}(z)>0\text{ and }\text{Im}(z)<0.
\displaystyle \text{Therefore, }z\text{ lies in the fourth quadrant.}
\displaystyle \arg(z)=-\beta=-\left(\frac{\pi}{2}-\alpha\right)=\alpha-\frac{\pi}{2}
\displaystyle \text{When }\alpha=0,\ z=-i\text{ and }\arg(z)=-\frac{\pi}{2}.
\displaystyle \therefore z=\sec\alpha\left\{\cos\left(\alpha-\frac{\pi}{2}\right)+i\sin\left(\alpha-\frac{\pi}{2}\right)\right\}.
\displaystyle \underline{\text{Case II: }}\alpha\in\left(\frac{\pi}{2},\pi\right]
\displaystyle z=\tan\alpha-i
\displaystyle |z|=\sqrt{\tan^2\alpha+1}
\displaystyle =\sqrt{\sec^2\alpha}=|\sec\alpha|=-\sec\alpha
\displaystyle \text{Let }\beta\text{ be the reference angle such that}
\displaystyle \tan\beta=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|.
\displaystyle \tan\beta=\left|\frac{-1}{\tan\alpha}\right|=|\cot\alpha|=-\cot\alpha
\displaystyle =\tan\left(\alpha-\frac{\pi}{2}\right)
\displaystyle \Rightarrow \beta=\alpha-\frac{\pi}{2}
\displaystyle \text{For }\frac{\pi}{2}<\alpha<\pi,\text{ we have }\text{Re}(z)<0\text{ and }\text{Im}(z)<0.
\displaystyle \text{Therefore, }z\text{ lies in the third quadrant.}
\displaystyle \arg(z)=-\pi+\beta
\displaystyle =-\pi+\alpha-\frac{\pi}{2}=\alpha-\frac{3\pi}{2}
\displaystyle \text{When }\alpha=\pi,\ z=-i\text{ and }\arg(z)=-\frac{\pi}{2}=\alpha-\frac{3\pi}{2}.
\displaystyle \therefore z=-\sec\alpha\left\{\cos\left(\alpha-\frac{3\pi}{2}\right)+i\sin\left(\alpha-\frac{3\pi}{2}\right)\right\}.
\displaystyle \\

\displaystyle \text{(iii) Let }z=(1-\sin\alpha)+i\cos\alpha.
\displaystyle \text{Both sine and cosine are periodic functions with period }2\pi.
\displaystyle \text{Hence, we may take }\alpha\in[0,2\pi].
\displaystyle |z|=\sqrt{(1-\sin\alpha)^2+\cos^2\alpha}
\displaystyle =\sqrt{2-2\sin\alpha}
\displaystyle =\sqrt{2}\sqrt{1-\sin\alpha}
\displaystyle =\sqrt{2}\sqrt{\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)^2}
\displaystyle =\sqrt{2}\left|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right|
\displaystyle \text{Let }\beta\text{ be the reference angle such that}
\displaystyle \tan\beta=\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|.
\displaystyle \tan\beta=\frac{|\cos\alpha|}{|1-\sin\alpha|}
\displaystyle =\left|\frac{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}}{\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)^2}\right|
\displaystyle =\left|\frac{\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}}\right|
\displaystyle =\left|\frac{1+\tan\frac{\alpha}{2}}{1-\tan\frac{\alpha}{2}}\right|
\displaystyle =\left|\tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right|
\displaystyle \underline{\text{Case I: }}0\leq\alpha<\frac{\pi}{2}
\displaystyle \cos\frac{\alpha}{2}>\sin\frac{\alpha}{2}
\displaystyle \text{and }\frac{\pi}{4}+\frac{\alpha}{2}\in\left[\frac{\pi}{4},\frac{\pi}{2}\right).
\displaystyle \therefore |z|=\sqrt{2}\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)
\displaystyle \tan\beta=\tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)
\displaystyle \Rightarrow \beta=\frac{\pi}{4}+\frac{\alpha}{2}
\displaystyle \text{Also, }\text{Re}(z)>0\text{ and }\text{Im}(z)>0.
\displaystyle \therefore z\text{ lies in the first quadrant and}
\displaystyle \arg(z)=\frac{\pi}{4}+\frac{\alpha}{2}.
\displaystyle \therefore z=\sqrt{2}\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)
\displaystyle \qquad\times\left\{\cos\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)+i\sin\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right\}.
\displaystyle \underline{\text{Case II: }}\frac{\pi}{2}<\alpha<\frac{3\pi}{2}
\displaystyle \cos\frac{\alpha}{2}<\sin\frac{\alpha}{2}
\displaystyle \text{and }\frac{\pi}{4}+\frac{\alpha}{2}\in\left(\frac{\pi}{2},\pi\right).
\displaystyle \therefore |z|=-\sqrt{2}\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)
\displaystyle \tan\beta=-\tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)
\displaystyle =\tan\left\{\pi-\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right\}
\displaystyle =\tan\left(\frac{3\pi}{4}-\frac{\alpha}{2}\right)
\displaystyle \Rightarrow \beta=\frac{3\pi}{4}-\frac{\alpha}{2}
\displaystyle \text{Also, }\text{Re}(z)>0\text{ and }\text{Im}(z)<0.
\displaystyle \therefore z\text{ lies in the fourth quadrant and}
\displaystyle \arg(z)=-\beta=\frac{\alpha}{2}-\frac{3\pi}{4}.
\displaystyle \therefore z=-\sqrt{2}\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)
\displaystyle \qquad\times\left\{\cos\left(\frac{\alpha}{2}-\frac{3\pi}{4}\right)+i\sin\left(\frac{\alpha}{2}-\frac{3\pi}{4}\right)\right\}.
\displaystyle \underline{\text{Case III: }}\frac{3\pi}{2}<\alpha\leq2\pi
\displaystyle \cos\frac{\alpha}{2}<\sin\frac{\alpha}{2}
\displaystyle \text{and }\frac{\pi}{4}+\frac{\alpha}{2}\in\left(\pi,\frac{5\pi}{4}\right].
\displaystyle \therefore |z|=-\sqrt{2}\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)
\displaystyle \tan\beta=\tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)
\displaystyle =\tan\left(\frac{\alpha}{2}-\frac{3\pi}{4}\right)
\displaystyle \Rightarrow \beta=\frac{\alpha}{2}-\frac{3\pi}{4}
\displaystyle \text{Also, }\text{Re}(z)>0\text{ and }\text{Im}(z)>0.
\displaystyle \therefore z\text{ lies in the first quadrant and}
\displaystyle \arg(z)=\frac{\alpha}{2}-\frac{3\pi}{4}.
\displaystyle \therefore z=-\sqrt{2}\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)
\displaystyle \qquad\times\left\{\cos\left(\frac{\alpha}{2}-\frac{3\pi}{4}\right)+i\sin\left(\frac{\alpha}{2}-\frac{3\pi}{4}\right)\right\}.
\displaystyle \text{When }\alpha=\frac{\pi}{2},\ z=0,\text{ so its argument and polar form are not defined.}
\displaystyle \text{When }\alpha=\frac{3\pi}{2},\ z=2=2(\cos0+i\sin0).
\displaystyle \\

\displaystyle \text{(iv) Let }z=\frac{1-i}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}.
\displaystyle 1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)
\displaystyle \text{Also, }\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\text{ has modulus }1\text{ and argument }\frac{\pi}{3}.
\displaystyle \therefore z=\frac{\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}
\displaystyle =\sqrt{2}\left[\cos\left(-\frac{\pi}{4}-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{4}-\frac{\pi}{3}\right)\right]
\displaystyle =\sqrt{2}\left[\cos\left(-\frac{7\pi}{12}\right)+i\sin\left(-\frac{7\pi}{12}\right)\right]
\displaystyle =\sqrt{2}\left(\cos\frac{7\pi}{12}-i\sin\frac{7\pi}{12}\right).
\displaystyle \therefore |z|=\sqrt{2}\text{ and }\arg(z)=-\frac{7\pi}{12}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }z_1\text{ and }z_2\text{ are two complex numbers such that }|z_1|=|z_2|
\displaystyle \text{and }\arg(z_1)+\arg(z_2)=\pi,\text{ show that }z_1=-\overline{z_2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta_1=\arg(z_1)\text{ and }\theta_2=\arg(z_2).
\displaystyle \text{Given }|z_1|=|z_2|\text{ and }\theta_1+\theta_2=\pi.
\displaystyle \therefore \theta_1=\pi-\theta_2.
\displaystyle \text{Now,}
\displaystyle z_1=|z_1|(\cos\theta_1+i\sin\theta_1)
\displaystyle =|z_2|\left[\cos(\pi-\theta_2)+i\sin(\pi-\theta_2)\right]
\displaystyle =|z_2|\left(-\cos\theta_2+i\sin\theta_2\right)
\displaystyle =-|z_2|\left(\cos\theta_2-i\sin\theta_2\right)
\displaystyle =-\overline{z_2}.
\displaystyle \therefore z_1=-\overline{z_2}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }z_1,z_2\text{ and }z_3,z_4\text{ are two pairs of conjugate complex}
\displaystyle \text{numbers, prove that }\arg\left(\frac{z_1}{z_4}\right)+\arg\left(\frac{z_2}{z_3}\right)=0.
\displaystyle \text{Answer:}
\displaystyle z_1,z_2\text{ and }z_3,z_4\text{ are two pairs of conjugate complex numbers.}
\displaystyle \therefore z_1=r_1e^{i\theta_1},\qquad z_2=r_1e^{-i\theta_1}
\displaystyle \text{and }z_3=r_2e^{i\theta_2},\qquad z_4=r_2e^{-i\theta_2}.
\displaystyle \therefore \frac{z_1}{z_4}=\frac{r_1e^{i\theta_1}}{r_2e^{-i\theta_2}}
\displaystyle =\frac{r_1}{r_2}e^{i(\theta_1+\theta_2)}
\displaystyle \Rightarrow \arg\left(\frac{z_1}{z_4}\right)=\theta_1+\theta_2\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }\frac{z_2}{z_3}=\frac{r_1e^{-i\theta_1}}{r_2e^{i\theta_2}}
\displaystyle =\frac{r_1}{r_2}e^{-i(\theta_1+\theta_2)}
\displaystyle \Rightarrow \arg\left(\frac{z_2}{z_3}\right)=-(\theta_1+\theta_2)\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle \arg\left(\frac{z_1}{z_4}\right)+\arg\left(\frac{z_2}{z_3}\right)
\displaystyle =(\theta_1+\theta_2)-(\theta_1+\theta_2)=0.
\displaystyle \therefore \arg\left(\frac{z_1}{z_4}\right)+\arg\left(\frac{z_2}{z_3}\right)=0.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Express }\sin\frac{\pi}{5}+i\left(1-\cos\frac{\pi}{5}\right)\text{ in polar form.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=\sin\frac{\pi}{5}+i\left(1-\cos\frac{\pi}{5}\right).
\displaystyle |z|=\sqrt{\sin^2\frac{\pi}{5}+\left(1-\cos\frac{\pi}{5}\right)^2}
\displaystyle =\sqrt{\sin^2\frac{\pi}{5}+1+\cos^2\frac{\pi}{5}-2\cos\frac{\pi}{5}}
\displaystyle =\sqrt{2\left(1-\cos\frac{\pi}{5}\right)}
\displaystyle =\sqrt{2\left(2\sin^2\frac{\pi}{10}\right)}
\displaystyle =2\sin\frac{\pi}{10}
\displaystyle \text{Let }\beta\text{ be the reference angle.}
\displaystyle \tan\beta=\frac{\left|1-\cos\frac{\pi}{5}\right|}{\left|\sin\frac{\pi}{5}\right|}
\displaystyle =\frac{2\sin^2\frac{\pi}{10}}{2\sin\frac{\pi}{10}\cos\frac{\pi}{10}}
\displaystyle =\tan\frac{\pi}{10}
\displaystyle \Rightarrow \beta=\frac{\pi}{10}
\displaystyle \text{Since }z\text{ lies in the first quadrant,}
\displaystyle \arg(z)=\frac{\pi}{10}.
\displaystyle \therefore \text{The polar form of }z\text{ is}
\displaystyle 2\sin\frac{\pi}{10}\left(\cos\frac{\pi}{10}+i\sin\frac{\pi}{10}\right).
\displaystyle \\

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