How to put $\frac{1}{\cos\theta - j\sin \theta}$ in the form $a+jb$? ($j^2=-1$) I am new to complex numbers and am having trouble putting them in the form $a+jb$ (or $a+bi$) How would I go about putting this expression in the form $a+jb$?
$$\frac{1}{\cos\theta - j\sin \theta}$$
 A: Hint. One may write, with $j^2=-1$,
$$ 
\frac{1}{\cos\theta - j\sin \theta}=\frac{1\cdot (\cos\theta+ j\sin \theta)}{(\cos\theta - j\sin \theta)(\cos\theta+ j\sin \theta)}.
 $$
A: $$\frac{1}{\cos\theta - j\sin \theta}=(\cos\theta - j\sin \theta)^{-1}=(e^{-\theta j})^{-1}=e^{\theta j}=\cos\theta + j\sin \theta$$
A: $$\frac{1}{\cos \theta - j \sin \theta}=\frac{\cos \theta + j \sin \theta}{(\cos \theta - j \sin \theta)(\cos \theta + j \sin \theta)}=\frac{\cos \theta+j \sin \theta}{\cos^{2}\theta-j^2 \sin^2 \theta}=\frac{\cos \theta + j \sin \theta}{\cos^2 \theta - (-1)\sin^2 \theta}=\frac{\cos \theta + j \sin \theta}{\cos^2 \theta + \sin^2 \theta}=\frac{\cos \theta + j \sin \theta}{1}=\cos \theta + j \sin \theta$$
A: Looks like you are not reading your math textbook . . . 
Conjugate pairings, e.g. $ (a + ib) $ and $ (a - ib) $, are the first thing explained after the definition of a complex number.
In basic algebra you would have seen that $$ a^2 - b^2 \equiv (a + b) (a - b) $$
So using this we can see that 
$$  (a + ib) (a - ib) \equiv a^2 - (ib)^2  \equiv a^2 + b^2 $$
So with any fraction with a complex number, we may eliminate the imaginary component below the division line by multiplying it by its conjugate as long as we also multiply the top part of the fraction by the conjugate also. 
You can now do the rest yourself.
