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If $z_1 = r_1(\cos\theta_1+i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2+i\sin\theta_2)$ prove that $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$ and that $\arg\left(\frac{z_1}{z_2}\right)=\arg z_1-\arg z_2$.

I've done most of this, but I got stuck at the end:

$$\begin{align} \frac{z_1}{z_2}&=\frac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2(\cos\theta_2+i\sin\theta_2)}\\ &=\frac{r_1}{r_2}\cdot\frac{(\cos\theta_1+i\sin\theta_1)(\cos\theta_2-i\sin\theta_2)}{1}\\ &=\frac{r_1}{r_2}\cdot[(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2)]\\ &=\frac{r_1}{r_2}\cdot[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)]\\ \\ \left|\frac{z_1}{z_2}\right|&=\sqrt{\frac{r_1^2}{r_2^2}\cdot[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)]^2} \end{align}$$ But in order for this to work, $[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)]^2$ must equal $1$ (so that $\left|\frac{z_1}{z_2}\right|=\frac{r_1}{r_2}$), but it doesn't. Where am I going wrong?

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  • $\begingroup$ Do you know that $|z_1z_2|=|z_1|\,|z_2|$ and that $\arg(z_1z_2)=\arg z_1+\arg z_2$? $\endgroup$
    – egreg
    Commented Nov 2, 2014 at 16:01
  • $\begingroup$ Where did that square root come from? Can't you use that |z| = r when z = r(cosθ+isinθ)? $\endgroup$ Commented Nov 2, 2014 at 16:02
  • $\begingroup$ @egreg When I look at the proof for that statement, it also ends up in the same situation: $|z_1 z_2|=\sqrt{r_1^2 r_2^2 [\cos(\theta_1+\theta_2+i\sin(\theta_1+\theta_2)]^2}=r_1 r_2$ $\endgroup$
    – hohner
    Commented Nov 2, 2014 at 16:03

4 Answers 4

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The step you take to the last equation isn't right; recall that the absolute value of a complex number is defined as $$|a+bi|=\sqrt{a^2+b^2}$$ but you've just written it as $$|z|=\sqrt{z^2}$$ which only works for real $z$. If you use the proper (first) identity, noting that the second to lasts equation is easy to separate into imaginary and real parts, you will get to $$\left|\frac{z_1}{z_2}\right|=\sqrt{\frac{r_1^2}{r_2^2}\cdot[\cos(\theta_1-\theta_2)^2+\sin(\theta_1-\theta_2)^2]}$$ which will work out as you expect.

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  • $\begingroup$ Ahh, I understand now. Thank you :) $\endgroup$
    – hohner
    Commented Nov 2, 2014 at 16:07
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If $z_1=r_1(\cos\theta_1+i\sin\theta_1)$ and $z_2=r_2(\cos\theta_2+i\sin\theta_2)$ it's just a matter of computations with the addition formulas to show that $$ z_1z_2=r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)) $$ so $$ |z_1z_2|=r_1r_2=|z_1|\,|z_2|,\qquad \arg(z_1z_2)=\theta_1+\theta_2=\arg z_1+\arg z_2 $$ For the quotient, set $w_1=z_1/z_2$ and $w_2=z_2$; then $$ |w_1w_2|=|w_1|\,|w_2| $$ or $$ |z_1|=\left|\frac{z_1}{z_2}\right|\,|z_2| $$ For the arguments it's the same.

In order to show that $|z_1z_2|=|z_1|\,|z_2|$ you can also prove first that $$ \overline{z_1z_2}=\overline{z_1}\,\overline{z_2} $$ which is just computations (use $z_1=a_1+ib_1$, $z_2=a_2+ib_2$). Then $$ |z_1z_2|^2=z_1z_2\cdot\overline{z_1z_2}= z_1z_2\overline{z_1}\overline{z_2}= z_1\overline{z_1}z_2\overline{z_2}= |z_1|^2|z_2|^2 $$ and the conclusion follows by taking square roots.

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Well $\frac{z_1}{z_2}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}$. And $|e^{i(\theta_1-\theta_2)}|=1$ So $|\frac{z_1}{z_2}|=\frac{r_1}{r_2}=\frac{|z_1|}{|z_2|}$ . all is used is Pythagorean's to get $|\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}|=1$

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Using the identity $|z|^2=z\overline{z}$, we have $$ \begin{aligned} \left|\frac{z_{1}}{z_{2}}\right| ^{2} &=\left(\frac{z_{1} }{z_{2}} \right)\overline{\left(\frac{z_{1}}{z_{2}}\right)}\\& =\left(\frac{z_{1}}{z_{2}}\right)\left(\frac{\bar{z}_{1}}{\bar{z}_{2}}\right)\\ &=\frac{z_{1} \bar{z}_{1}}{z_{2} \bar{z}_{2}} \\ &=\frac{\left|z_{1}\right|^{2}}{\left|z_{2}\right|^{2}} \\ &=\left(\frac{|z_{1} \mid}{\left|z_{2}\right|}\right)^{2} \end{aligned} $$

We can conclude that $$ \left|\frac{z_{1}}{z_{2}}\right|=\frac{\left|z_{1}\right|}{\left|z_{2}\right|} $$

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