We have:

$z_1 = 2 + 2i$ and $z_2 = -1 -\sqrt{3} i$

I am asked for obtaining arg$\left(\frac{z_1}{z_2}\right)$ and arg($z_1z_2$)

As we know:

$$\theta = \tan^{-1} \left(\frac{y}{x}\right)$$

The division of complex numbers is:

$$\frac{z_1}{z_2} = z_1z_2^{-1} = \frac{z_1z_2^*}{z_2z_2^*}$$

Doing so I got $\frac{\sqrt{3}-1}{-1-\sqrt{3}}$ but I do not know how to get the angle without calculator using this result.

The same happened to me in the product:

$$arg(z_1z_2) = \tan^{-1}\left(\frac{-1-\sqrt{3}}{-1+\sqrt{3}}\right) = ...$$

I know it has to be a silly thing but I am stuck here.


You can try proving by yourself that:

$$ \arg(z_1z_2)=\arg(z_1)+\arg(z_2) $$

Which by extension yields: $$ \arg(z_1/z_2)=\arg(z_1)-\arg(z_2) $$

In your case: $$ \arg(z_1/z_2)=45°-230°=-185° $$

  • 1
    $\begingroup$ Usually, argument should lie between $\left( - \pi, \pi \right]$ and hence cannot be $-185^o$. $\endgroup$ – Aniruddha Deshmukh Aug 20 '18 at 19:17

I do not really know if you wanted this. But, try using the following steps to find the value

$$tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{- 1 + \sqrt{3}} \right) = \tan^{-1} \left( \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} \right)$$ $$\therefore \tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{-1 + \sqrt{3}} \right) = \tan^{-1} \left( \dfrac{\tan \left( \dfrac{\pi}{4} \right) + \tan \left( \dfrac{\pi}{3} \right)}{1 - \tan \left( \dfrac{\pi}{4} \right) \cdot \tan \left( \dfrac{\pi}{3} \right)} \right)$$ $$\therefore \tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{-1 + \sqrt{3}} \right) = \tan^{-1} \left( \tan \left( \dfrac{7 \pi}{12} \right) \right) = \dfrac{7 \pi}{12}$$

You can try the same method for other as well.


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