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A question in my textbook asks:

Find $\frac{z_1}{z_2}$ if $z_1=2\left(\cos\left(\frac{\pi}3\right)+i\sin\left(\frac{\pi}3\right)\right)$ and $z_2=\cos\left(\frac{\pi}6\right)-i\sin\left(\frac{\pi}6\right)$. I converted $z_2$ to $\cos\left(-\frac{\pi}6\right)+i\sin\left(-\frac{\pi}6\right)$ as I initially thought it would be easier to use Euler's identity (which it is) but the textbook hadn't introduced this yet so it must be possible without having to use it. As a result, I am stuck at square one, any help would be great. Thanks.

Shahar
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user175978
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3 Answers3

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Just an expansion of my comment above: presumably you know how to do $$ \alpha(a+bi)(c+di)\quad\text{here}\quad i=\sqrt{-1}; a,b,c,d,\alpha\in\mathbb{R}. $$ Then for $c+di\neq 0$, we have $$ \frac{a+bi}{c+di}=\alpha(a+bi)(c-di)\quad\text{with}\quad\alpha=\frac{1}{c^2+d^2}. $$ In fact, this is usually how we define division by a nonzero complex number.

In your case, $a,b,c$ and $d$ are all given so just plug in the numbers. And with $a,b,c$ and $d$ being trig functions, I'm sure some simplication is going to happen.

Kim Jong Un
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If you are working with complex number in the form you gave, recall that $r\cos\theta+ir\sin\theta=re^{i\theta}$.

You then multiply and divide complex numbers in polar form in the natural way:

$$r_1e^{1\theta_1}\cdot r_2e^{1\theta_2}=r_1r_2e^{i(\theta_1+\theta_2)},$$

$$\frac{r_1e^{1\theta_1}}{r_2e^{1\theta_2}}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}$$

rotten
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  • How would I do it without using the natural way (i.e using the trigonometrical functions) the textbook hadn't introduced that identity at this point so it must be possible. – user175978 Sep 14 '14 at 15:49
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$$z_{1}=2(cos(\frac{pi}{3})+i sin (\frac{pi}{3}) )=2e^{i\frac{pi}{3}}\\z_{2}=1(cos(\frac{pi}{6})-i sin (\frac{pi}{6}) )=1(cos(\frac{pi}{6}) +i sin (\frac{-pi}{6}) )=\\as-we-know\\cos(a)=cos(-a)\\1(cos(\frac{-pi}{6})-i sin (\frac{-pi}{6}) )=1e^{\frac{-pi}{6}\\ $$

Khosrotash
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