Question

For any complex numbers $z,z_1$ and $z_2$ prove that:
(i) $arg\left(\overline{z}\right)=-arg\left(z\right)$
(ii) $arg\left(z_1{z}_2\right)=arg\left(z_1\right)+arg\left(z_2\right)$
(iii) $arg\left(z_1\overline{z_2}\right)=arg\left(z_1\right)-arg\left(z_2\right)$
(iv) $arg\left(\frac{z_1}{z_2}\right)=arg\left(z_1\right)-arg\left(z_2\right)$

Answer 1

$\left(i\right)\text{Let}z=r(\cos \theta +i\sin \theta ).\text{Then},|z|=r\text{and arg}\left(z\right)=\theta .\text{Now,}z=r(\cos \theta +i\sin \theta )\text{Complex numbers and quadratic Equations}\Rightarrow z=r\cos \theta +i\left(r\sin \theta \right)\Rightarrow \overline{z}=r\cos \theta =i\left(r\sin \theta \right)=r(cos\theta -i\sin \theta )=r\left\{\cos \right(-\theta )+i\sin (-\theta \left)\right\}\text{hence, arg}\left(\overline{z}\right)=-arg\left(z\right).\left(ii\right)\text{Let}{z}_1=r_1(\cos {\theta }_1+is\sin {\theta }_1)\text{and}{z}_2=r_2(\cos {\theta }_2+i\sin {\theta }_2).\text{then}|z_1|=r_1,\text{arg}\left(z_1\right)=\theta \text{and}|z_2|=r_2,arg\left(z_2\right)={\theta }_2.\therefore z_1{z}_2=r_1(\cos {\theta }_1+i\sin {\theta }_1).r_2(\cos {\theta }_2+i\sin {\theta }_2)=r_1{r}_2\left\{\right(\cos {\theta }_1\cos {\theta }_1-\sin {\theta }_1\sin {\theta }_2)+i(\sin {\theta }_1\cos {\theta }_2+\cos {\theta }_2+\cos {\theta }_1\sin {\theta }_2\left)\right\}=r_1{r}_2\left\{\cos \right({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2\left)\right\}\Rightarrow arg\left(z_1{z}_2\right)=({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2).\text{REMARKS}\left(\right(I\left)\text{Note here that}\right)|z_1{z}_2|=r_1{r}_2=|z_1||z_2|.\text{(II) In general, we have}|z_1{z}_2\cdots z_n|=|z_1|.|z_2|\cdots |z_n|\text{Let arg}(z_1{z}_2\cdots z_n)=arg\left(z_1\right)+\text{arg}\left(z_2\right)+\cdots +\text{arg}\left(z_n\right).\text{(iii) Let}{z}_1=r_1(\cos {\theta }_1+i\sin {\theta }_1)and{z}_2=r_2(cos{\theta }_2+i\sin {\theta }_2).\text{then,}{\overline{z}}_2=\overline{r_2\cos \theta +i\left(r_2\sin {\theta }_2\right)}=r_2\cos {\theta }_2-i\left(r_2\sin {\theta }_2\right)\Rightarrow {\overline{z}}_2=r_2\left\{\cos \right(-{\theta }_2)+i\sin (-{\theta }_2\left)\right\}.\therefore z_1{\overline{z}}_2=r_1(\cos {\theta }_1+i\sin {\theta }_1).r_2\left\{\cos \right(-{\theta }_2)+i\sin (-{\theta }_2\left)\right\}=r_1{r}_2(\cos {\theta }_1+i\sin {\theta }_1)\left\{\cos \right(-{\theta }_2)+i\sin \}(-{\theta }_2)=r_1{r}_2\left[\cos \right\{{\theta }_1+(-{\theta }_2)\}+i\sin \{{\theta }_1+(-{\theta }_2)\left\}\right]=r_1{r}_2\left\{\cos \right({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2\left)\right\}.\text{Hence, arg}\left(z_1{\overline{z}}_2\right)=({\theta }_1-{\theta }_2)+=arg\left(z_1\right)-arg\left(z_2\right)\text{Let}{z}_1=r_1(\cos {\theta }_1+i\sin {\theta }_1)\text{and}{z}_2=r_2(\cos {\theta }_2+i\sin {\theta }_2)\text{then,}|z_1|=r_1,|z_2|=r_2,arg\left(z_1\right)={\theta }_1\text{and arg}\left(z_2\right)={\theta }_2.\therefore \frac{z_1}{z_2}=\frac{r_1(\cos {\theta }_1+i\sin {\theta }_1)}{r_2(\cos {\theta }_2+i\sin {\theta }_2)}\times \frac{(\cos {\theta }_2-i\sin {\theta }_2)}{(\cos {\theta }_2-i\sin {\theta }_2)}=\frac{r_1}{r_2}.\left\{\frac{(\cos {\theta }_1.\cos {\theta }_2-\sin {\theta }_1.\sin {\theta }_2)+i(\sin {\theta }_1.\cos {\theta }_2-\cos {\theta }_1.\sin {\theta }_2)}{({\cos }^2{\theta }_2+{\sin }^2{\theta }_2)}\right\}=\frac{r_1}{r_2}.\left\{\cos \right({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2\left)\right\}\Rightarrow arg\left(\frac{z_1}{z_2}\right)=({\theta }_1-{\theta }_2)=arg\left(z_1\right)-arg(z_2.)\text{Hence, arg}\left(\frac{z_1}{z_2}\right)=\arg \left(z_1\right)-arg\left(z_2\right)$