# If $z=\cos \theta + i \sin \theta$, express $\displaystyle \frac {1}{1-z \cos \theta}$ in the form $a+i\cdot b$.

In order to get rid of the $$z$$, should I substitute $$z=\cos\theta + i\sin \theta$$ into the complex number, or what conjugate should I multiply the complex number by?

I have tried substituting $$z=\cos\theta + i\sin \theta$$ into the complex number, but only got this far:

$$\displaystyle \frac {1}{1-z \cos \theta}$$

$$= \displaystyle\frac{1}{1-(\cos\theta + i\sin\theta)(\cos\theta)}$$

$$=\displaystyle\frac{1}{1-\cos^2\theta-i\cos\theta\sin\theta}$$

As for multiplying the complex number by a conjugate, I have used $$\big(\displaystyle\frac {1}{z}-\frac{1}{\cos\theta}\big)$$, $$\big(\displaystyle z-\frac{1}{z}\big)$$ and $$(1+z\cos\theta)$$ but to no avail.

I have only learnt de Moivre's theorem, and I haven't learnt $$\cosθ+i\sinθ=e^{iθ}$$, so I would appreciate if this question can be solved in the simplest way possible. But other methods are welcome.

$$\dfrac1{1-\cos\theta(\cos\theta+i\sin\theta)}=\dfrac1{\sin^2\theta-i\sin\theta\cos\theta}=\dfrac1{-i\sin\theta}\cdot\dfrac1{\cos\theta+i\sin\theta}=\dfrac{\cos\theta-i\sin\theta}{-i\sin\theta}$$
• from your last step $\displaystyle\frac{\cos\theta + i \sin\theta}{-i\sin\theta}$, I continued the working but got stuck at getting the final answer. \begin{align} \frac{\cos\theta + i \sin\theta}{-i\sin\theta} &= \frac{i\sin\theta}{-i\sin\theta}+\frac{\cos\theta}{-i\sin\theta} \\&= -1-\frac{\cot\theta}{i}\end{align} the answer is $1+i\cot\theta$. How do I get it? Mar 23 '20 at 5:31
\begin{align} \frac1{1-z\cos\theta}&=\frac1{1-(\cos\theta+i\sin\theta)\cos\theta}\\ &=\frac1{1-\cos^2\theta-i\sin\theta\cos\theta}\\ &=\frac1{\sin^2\theta-i\sin\theta\cos\theta}\color{blue}{\cdot\frac{\sin^2\theta+i\sin\theta\cos\theta}{\sin^2\theta+i\sin\theta\cos\theta}}\\ &=\frac{\sin^2\theta+i\sin\theta\cos\theta}{\sin^4\theta+\sin^2\theta\cos^2\theta}\\ &=\frac{\sin^2\theta+i\sin\theta\cos\theta}{\sin^2\theta(\sin^2\theta+\cos^2\theta)}\\ &=\frac{\sin^2\theta}{\sin^2\theta}+\frac{i\sin\theta\cos\theta}{\sin^2\theta}\\ &=1+i\cot\theta \end{align}
Note that $$z=e^{i\theta}$$, $$z=e^{-i\theta}$$ and $$\cos\theta = \frac{z+\bar z}2$$
$$\frac {1}{1-z \cos \theta}=\frac {1}{1-z \cos \theta}\cdot\frac {1-\bar z \cos \theta}{1-\bar z \cos \theta}$$ $$=\frac {1-e^{-i\theta}\cos \theta}{1-(\bar z +z)\cos \theta + \cos^2\theta}$$ $$=\frac {1-(\cos\theta - i\sin\theta)\cos \theta}{1-2\cos\theta\cos \theta + \cos^2\theta}$$ $$=\frac {1-\cos^2\theta + i\sin\theta\cos \theta}{1- \cos^2\theta}$$ $$=1+ i\cot\theta$$