First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.5

Solution

Let:

z_1=r_1(\cos\theta_1+i\sin\theta_1),\quad z_2=r_2(\cos\theta_2+i\sin\theta_2)

(i) $\operatorname{Arg}(z_1z_2)=\operatorname{Arg}(z_1)+\operatorname{Arg}(z_2)$

\begin{aligned} z_1z_2 &=r_1r_2(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)\\ &=r_1r_2\big[(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2) +i(\cos\theta_1\sin\theta_2+\sin\theta_1\cos\theta_2)\big]\\ &=r_1r_2\big[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\big] \end{aligned}

So:

\boxed{\operatorname{Arg}(z_1z_2)=\theta_1+\theta_2=\operatorname{Arg}(z_1)+\operatorname{Arg}(z_2)}

(ii) $\operatorname{Arg}\left(\dfrac{z_1}{z_2}\right)=\operatorname{Arg}(z_1)-\operatorname{Arg}(z_2)$

\begin{aligned} \frac{z_1}{z_2} &=\frac{r_1}{r_2}\cdot\frac{\cos\theta_1+i\sin\theta_1}{\cos\theta_2+i\sin\theta_2}\\ &=\frac{r_1}{r_2}(\cos\theta_1+i\sin\theta_1)(\cos\theta_2-i\sin\theta_2)\\ &=\frac{r_1}{r_2}\big[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\big] \end{aligned}

Therefore:

\boxed{\operatorname{Arg}\left(\frac{z_1}{z_2}\right)=\theta_1-\theta_2=\operatorname{Arg}(z_1)-\operatorname{Arg}(z_2)}

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