# How would I prove $\frac{1}z=\operatorname{cis}(-\theta)$ given $z=\operatorname{cis}(\theta)$?

I know $$\operatorname{cis}(-\theta)$$ equals $$\cos(\theta)-i\sin(\theta)$$ and $$\frac{1}z=\frac{1}\cos+i\frac{1}\sin=\sec+\frac{\csc}i$$ (please, correct me if I'm wrong.)

I just don't know how to relate these two now. Can anyone provide some guidance?

• Wrote $cis(-\theta) =e^{-i\theta}$ to reach answer – Rohan Shinde Mar 11 '19 at 3:11
• what do you mean exactly? – Ryan_DS Mar 11 '19 at 3:12
• If $z=e^{i\theta}=\cos \theta + i \sin \theta = e^{i\theta}$ then $\frac 1 z =e^{-i\theta}=\cos\theta-i\sin\theta$ – J. W. Tanner Mar 11 '19 at 3:16

## 2 Answers

assuming you mean $$\text{cis}(\theta)=e^{\theta i}=\cos\theta+i\sin\theta$$ you can just write $$\text{cis}(-\theta)=e^{-\theta i}=\cos(-\theta)+i\sin(-\theta)=\cos\theta-i\sin\theta\\ \frac{1}{z}=\frac{1}{\cos\theta+i\sin\theta}\\ =\frac{1}{\cos\theta+i\sin\theta}\cdot\frac{\cos\theta-i\sin\theta}{\cos\theta-i\sin\theta}\\ =\frac{\cos\theta-i\sin\theta}{\cos^2\theta-(i\sin\theta)^2}\\ =\frac{\cos\theta-i\sin\theta}{\cos^2\theta+\sin^2\theta}=\cos\theta-i\sin\theta=\text{cis}(-\theta)$$ where you just multiply by the conjugate to eliminate the complex number on below part, and use the facts that $$i^2=-1$$ and $$\cos^2\theta+\sin^2\theta=1.$$

P.s.: the step you took

$$\frac{1}{z}=\frac{1}{\cos\theta}+\frac{i}{\sin\theta}$$

is wrong because you can't generally say that $$\frac{1}{a+bi}=\frac{1}{a}+\frac{i}{b}.$$

It is not correct in general to say if $$z=a+bi$$ then $$\frac 1 z = \frac 1 a + \frac 1 b i;$$

rather, $$\frac 1 z = \frac a {a^2+b^2} - \frac b {a^2+b^2}i.$$

Note that $$\cos^2\theta + \sin^2\theta = 1$$.

Therefore, if $$z=e^{i\theta}=\cos \theta + i \sin \theta$$ then $$\frac 1 z =e^{-i\theta}=\cos\theta-i\sin\theta.$$

• You could also say $e^{-i\theta}=\cos(-\theta)+i\sin(-\theta)=\cos(\theta)-i\sin(\theta)$ – J. W. Tanner Mar 11 '19 at 3:27
• The formula I gave for $\frac 1 z$ in terms of $a$ and $b$ assumes $z=a+bi\ne0$, which holds when $z=e^{i\theta}$ – J. W. Tanner Mar 11 '19 at 3:30
• i don't think the division is defined for z=0, not like that the case since $e^{i\theta}$ is clearly non nule :v. – cand Mar 11 '19 at 3:33
• @cand: that's what I was trying to say – J. W. Tanner Mar 11 '19 at 3:34