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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?

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

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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 $\frac 1 z =e^{-i\theta}=\cos\theta-i\sin\theta.$

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    $\begingroup$ You could also say $e^{-i\theta}=\cos(-\theta)+i\sin(-\theta)=\cos(\theta)-i\sin(\theta)$ $\endgroup$ – J. W. Tanner Mar 11 '19 at 3:27
  • $\begingroup$ The formula I gave for $\frac 1 z$ in terms of $a$ and $b$ assumes $z=a+bi\ne0$, which holds when $z=e^{i\theta}$ $\endgroup$ – J. W. Tanner Mar 11 '19 at 3:30
  • $\begingroup$ 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. $\endgroup$ – cand Mar 11 '19 at 3:33
  • $\begingroup$ @cand: that's what I was trying to say $\endgroup$ – J. W. Tanner Mar 11 '19 at 3:34

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