Prop${}^s$ of Arg

• $ \arg (\bar{z})=-\arg z$

$\quad \mathrm{z}=\mathrm{x}+\mathrm{iy}$

$\quad =\mathrm{r}(\cos \theta+ i\sin \theta) ………$ polar form

$\quad =\mathrm{re}^{\mathrm{i} \theta} ………$ Euler’s form

$\quad \arg z=\theta$

$\quad \overline{\mathrm{z}}=\mathrm{r}(\cos \theta-i \sin \theta)$

$\quad =\mathrm{re}^{-\mathrm{i} \theta}$

$\quad =\mathrm{r}\cos (-\theta)+\mathrm{isin}(-\theta)$

$\quad \arg \overline{\mathrm{z}}=-\theta=-\arg \mathrm{z}$.

• $ \arg \left(z _{1} z _{2}\right)=\arg z _{1}+\arg z _{2}$

$\quad \mathrm{z} _{1}=\mathrm{r} _{1}\left(\cos \theta _{1}+\mathrm{isin} \theta _{1}\right) \quad \arg \mathrm{z} _{1}=\theta _{1}$

$\quad \mathrm{z} _{2}=\mathrm{r} _{2}\left(\cos \theta _{2}+\mathrm{isin} \theta _{2}\right) \quad \arg \mathrm{z} _{2}=\theta _{2}$

$\quad \mathrm{z} _{1} \mathrm{z} _{2}=\mathrm{r} _{1} \mathrm{r} _{2} e^{i\left(\theta _{1}+\theta _{2}\right)}$

$\quad =\mathrm{r}_1 \mathrm{r}_2\left[\cos \left[\left(\theta_1+\theta_2\right)+i \sin \left(\theta_1+\theta_2\right)\right]\right.$

$\quad \arg \left(z _{1} z _{2}\right)=\theta _{1}+\theta _{2}$

$\quad =\arg z _{1}+\arg z _{2}$

$\quad$ In general, we write $\arg \left(z _{1} z _{2}\right)=\arg z _{1}+\arg z _{2}+2 k \pi$, where k is $- l$ or $0$ or $1$

• $\quad \arg \left(\frac{z _{1}}{z _{2}}\right)=\arg \left(z _{1}\right)-\arg z _{2}+2 k \pi$ where $\mathrm{k}=0,-1,1$

$\quad \left(\frac{\mathrm{z} _{1}}{\mathrm{z} _{2}}\right)=\left(\frac{\mathrm{r} _{1}}{\mathrm{r} _{2}}\right) \mathrm{e}^{\mathrm{i}\left(\theta _{1}-\theta _{2}\right)} $

$\quad =\left(\frac{\mathrm{r} _{1}}{\mathrm{r} _{2}}\right)\left[\cos \left(\theta _{1}-\theta _{2}\right)+i \sin \left(\theta _{1}-\theta _{2}\right)\right]$

$\quad \arg \left(\frac{\mathrm{z} _{1}}{\mathrm{z} _{2}}\right)=\theta _{1}-\theta _{2}$

$\quad =\arg z _{1}-\arg z _{2}$

• $\quad \arg \left(\mathrm{z}^{\mathrm{n}}\right)=\mathrm{n} \arg \mathrm{z}$

$\quad \mathrm{z}=\mathrm{r}(\cos \theta+\mathrm{isin} \theta)$

$\quad \mathrm{z}=\mathrm{r} e^{i \theta}$

$\quad \mathrm{z}^{\mathrm{n}}=\mathrm{r}^{\mathrm{n}} \mathrm{e}^{\mathrm{in} \theta}$

$\quad \mathrm{Z}^{\mathrm{n}}=\mathrm{r}^{\mathrm{n}}[\cos (\mathrm{n} \theta)+\mathrm{isinn} \theta]$

$\quad \arg \mathrm{z}^{\mathrm{n}}=\mathrm{n} \theta=\mathrm{n} \arg \mathrm{z}$

Que: 1 If $\arg z _{1}=170^{\circ}$ and $\arg z _{2}=70^{\circ}$, then final the principal argument of $z _{1} z _{2}$ $\arg \left(z _{1} z _{2}\right)=\arg \left(z _{1}\right)+\arg z _{2}=240^{\circ}$

Thus $z _{1} z _{2}$ lies in $3^{\text {rd }}$ quadrant. Hence its principal argument is $-120^{\circ}$.

Que: 2 If $z _{1}$ and $z _{2}$ are conjugate to each other, then find $\arg \left(-z _{1} z _{2}\right)$.

$\because \mathrm{z} _{1} \& \mathrm{z} _{2}$ are conjugate to each other i.e $\mathrm{z} _{2}=\overline{\mathrm{z}} _{1}$

$\therefore \arg \left(-\mathrm{z} _{1} \mathrm{z} _{2}\right)=\arg \left(-\mathrm{z} _{1} \overline{\mathrm{z}} _{1}\right)=\arg \left(-\left|\mathrm{z} _{1}\right|^{2}\right)$

$=\arg$ (negative real no.)

$=\pi$.

Que: 3 If $z _{1}, z _{2} \& z _{3}, z _{4}$ are two pairs of conjugate complex nos, then find the value of $\arg \left(\frac{z _{1}}{z _{4}}\right)+\arg \left(\frac{z _{2}}{z _{3}}\right)$. $z _{2}=\bar{z} _{1} \& z _{4}=\bar{z} _{3}$ therefore $z _{1} z _{2}=\left|z _{1}\right|^{2} \& z _{3} z _{4}=\left|z _{3}\right|^{2}$

Now $\arg \left(\frac{z _{1}}{z _{4}}\right)+\arg \left(\frac{z _{2}}{z _{3}}\right)=\arg \left(\frac{z _{1} z _{2}}{z _{3} z _{4}}\right)=\arg \left(\frac{\left|z _{1}\right|^{2}}{\left|z _{3}\right|^{2}}\right)=\arg \left(\left|\frac{z _{1}}{\left.z _{3}\right|^{2}}\right|^{2}\right)=\arg$ (positive real no.) $=0$

Properties of Modulus

$1 \left|z _{1} z _{2}\right|=\left|z _{1}\right|\left|z _{2}\right|$

Explanation: Let $\mathrm{z} _{1}=\mathrm{x} _{1}+\mathrm{iy} _{1} \& \mathrm{z} _{2}=\mathrm{x} _{2}+\mathrm{iy} _{2}$

Now $\left|\mathrm{z} _{1} \mathrm{z} _{2}=\left(\mathrm{x} _{1}+\mathrm{iy} _{1}\right)\left(\mathrm{x} _{2}+\mathrm{iy} \mathrm{y} _{2}\right)=\left(\mathrm{x} _{1} \mathrm{x} _{2}-\mathrm{y} _{1} \mathrm{y} _{2}\right)+\mathrm{i}\left(\mathrm{x} _{1} \mathrm{y} _{2}+\mathrm{x} _{2} \mathrm{y} _{1}\right)\right|=\sqrt{\left(x _{1} x _{2}-y _{1} y _{2}\right)^{2}+\left(x _{1} x _{2}+y _{1} y _{2}\right)^{2}}$

$=\sqrt{\left(x _{1}^{2}+y _{1}^{2}\right)\left(x _{2}^{2}+y _{2}^{2}\right)}=\sqrt{\left(x _{1}^{2}+y _{1}^{2}\right)} \sqrt{\left(x _{2}^{2}+y _{2}^{2}\right)}=\left|z _{1}\right|\left|z _{2}\right|$

as $\left|\mathrm{z} _{1} \mathrm{z} _{2} \ldots \ldots \ldots \ldots \ldots \ldots \mathrm{z} _{\mathrm{n}}\right|=\left|\mathrm{z} _{1}\right|\left|\mathrm{z} _{2}\right| \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .\left|\mathrm{z} _{\mathrm{n}}\right|$

so, $\left|z^{n}\right|=\mid z . z . \ldots \ldots \ldots \ldots \ldots$. .to $n$ factors $|=| z|z| \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .$. .to $n$ factors $=|z|^{n}$.

$2 \quad\left|\frac{\mathrm{z} _{1}}{\mathrm{z} _{2}}\right|=\frac{\mathrm{z} _{1}}{\mathrm{z} _{2}}$ (try yourself)

$3 \quad\left|z _{1}+z _{2}\right|^{2}=\left|z _{1}\right|^{2}+\left|z _{2}\right|^{2}+2 \operatorname{re}\left(z \bar{z} _{2}\right)$

$4 \quad\left|z _{1}-z _{2}\right|^{2}=\left|z _{1}\right|^{2}+\left|z _{2}\right|^{2}-2 \operatorname{re}\left(z \bar{z} _{2}\right)$

Explanation of (3) &(4): $\because z \bar{z}=|z|^{2}$

$\therefore\left|\mathrm{Z}_1 \pm \mathrm{Z}_2\right|^2=\left(\mathrm{Z}_1{ }^{ \pm} \mathrm{Z}_2\right)\left(\overline{\mathrm{Z}_1{ }^{ \pm} \mathrm{Z}_2}\right)=\left(\left(Z_1{ }^{ \pm} Z_2\right)\left(\overline{Z_1}{ }^{ \pm} \overline{Z_2}\right)\right)={Z_1} \bar{Z}_1+Z_2 \overline{Z_2}{ }^{ \pm} Z_1 \overline{Z_2}{ }^{ \pm} Z_2 \overline{Z_1}$

$\Rightarrow\left|z _{1}+z _{2}\right|^{2}=\left|z _{1}\right|^{2}+\left|z _{2}\right|^{2} \pm 2 \operatorname{re}\left(z _{1} \bar{z} _{2}\right)\left(\overline{z _{1} \bar{z} _{2}}=\overline{z _{1} z _{2}}=\overline{z _{1}} z _{2}\right)$

$5 \quad\left|z _{1}+z _{2}\right|^{2}+\left|z _{1}-z _{2}\right|^{2}=2\left(\left|z _{1}\right|^{2}+\left|z _{2}\right|^{2}\right)$

Explanation: Adding (3) & (4), we get the result.

$6 \quad\left|z _{1}+z _{2}\right| \leq\left|z _{1}\right|+\left|z _{2}\right|$ (Triangle Inequality)

Explanation: $\left|z _{1}+z _{2}\right|=\mid r _{1}\left(\cos \theta _{1}+i \sin \theta _{1}\right)+r _{2}\left(\cos \theta _{2}+i \sin \theta _{2} \mid\right.$

$=\left(r _{1} \cos \theta _{1}+r _{2} \cos \theta _{2}\right)+i\left(r _{1} \sin \theta _{1}+r _{2} \sin \theta _{2}\right)=$

$=\sqrt{\left(r _{1} \cos \theta _{1}+r _{2} \cos \theta _{2}\right)^{2}+\left(r _{1} \sin \theta _{1}+r _{2} \sin \theta _{2}\right)^{2}}$

$=\sqrt{r _{1}{ }^{2}\left(\cos ^{2} \theta _{1}+\sin ^{2} \theta _{1}\right)+r _{2}{ }^{2}\left(\cos ^{2} \theta _{2}+\sin ^{2} \theta _{2}\right)+2 r _{1} r _{2}\left(\cos \theta _{1} \cos \theta _{2}+\sin \theta _{1} \sin \theta _{2}\right)}$

$=\sqrt{r _{1}^{2}+r _{2}^{2}+2 r _{1} r _{2} \cos \left(\theta _{1}-\theta _{2}\right)}$

$\leq \sqrt{r _{1}^{2}+r _{2}^{2}+2 r _{1} r _{2}}\left(\because \cos \left(\theta _{1}-\theta _{2}\right) \leq 1\right)$

$\therefore\left|\mathrm{z} _{1}+\mathrm{z} _{2}\right| \leq \sqrt{\left(r _{1}+r _{2}\right)^{2}}$

or $\therefore\left|\mathrm{z} _{1}+\mathrm{z} _{2}\right| \leq r _{1}+r _{2}$

Thus $\left|z _{1}+z _{2}\right| \leq\left|z _{1}\right|+\left|z _{2}\right|$

$7 \quad\left|z _{1}-z _{2}\right| \geq|| z _{1}|-| z _{2} \mid$

Explanation: $\left|z_1-z_2\right|=\mid r_1\left(\cos \theta_1+i \sin \theta_1\right)-r_2\left(\cos \theta_{2_1}+i \sin \theta_2 \mid\right)$

$=\left(r _{1} \cos \theta _{1}-r _{2} \cos \theta _{2}\right)+i\left(r _{1} \sin \theta _{1}-r _{2}{ }^{\sin } \theta _{2}\right)=$

$=\sqrt{\left(r _{1} \cos \theta _{1}-r _{2} \cos \theta _{2}\right)^{2}+\left(r _{1} \sin \theta _{1}-r _{2} \sin \theta _{2}\right)^{2}}$

$=\sqrt{r _{1}^{2}\left(\cos ^{2} \theta _{1}+\sin ^{2} \theta _{1}\right)+r _{2}^{2}\left(\cos ^{2} \theta _{2}+\sin ^{2} \theta _{2}\right)-2 r _{1} r _{2}\left(\cos \theta _{1} \cos \theta _{2}+\sin \theta _{1} \sin \theta _{2}\right)}$

$=\sqrt{r _{1}^{2}+r _{2}^{2}-2 r _{1} r _{2} \cos \left(\theta _{1}-\theta _{2}\right)}$

$\geq \sqrt{r _{1}^{2}+r _{2}^{2}-2 r _{1} r _{2}}\left(\because \cos \left(\theta _{1}-\theta _{2}\right) \leq 1\right)$

$\therefore\left|\mathrm{z} _{1}-\mathrm{z} _{2}\right| \geq \sqrt{\left(r _{1}-r _{2}\right)^{2}}$

or $\therefore\left|z _{1}-z _{2}\right| \geq\left|r _{1}-r _{2}\right|=\left|z _{1}\right|-\left|z _{2}\right|$

Thus $\left|z _{1}-z _{2} \geq\right| z _{1}|-| z _{2} \mid$