Complex numbers trig form

Question

Given that $z= \cos(x) + i\sin(x),$ show that $\frac{1}{1+z}= 1 + i\tan(x/2)$ where x is not equal to pi/2 I tried to add 1 to z and then invert and realize the denominator but didn't get anywhere with that

Answer 1

$z=\cos x+i\sin x=e^{ix}$ and $$\tan(x/2)=\frac{e^{ix/2}-e^{-ix/2}}{i(e^{ix/2}+e^{-ix/2})}=\frac{e^{ix}-1}{i(e^{ix}+1)}=\frac{z-1}{i(z+1)}$$ So $$i\tan(x/2)=i\frac{z-1}{i(z+1)}=\frac{z-1}{z+1}=1-\frac{2}{z+1}$$

You can see your equality is wrong, but it could be $$\frac1{1+z}=\frac12(1-i\tan(x/2))$$

Answer 2

Your equality does not hold for $x=\pi/2$, since $\cos\frac\pi2 = 0$ and $\sin\frac\pi2 = 1$, meaning $z=0 + i\cdot 1 = i$

$$\frac{1}{1+z} = \frac{1}{1+i}=\frac{1-i}{(1+i)(1-i)}=\frac{1-i}{2},$$

while $\tan\frac{x}{2} = \tan\frac{\pi}{4} = 1$, meaning $$1+i\tan\left(\frac x2\right) = 1+i$$

Answer 3

$$\begin{align}\frac{1}{1+\cos x + i\sin x} &= \frac{1}{1+\Big(2\cos ^2 \dfrac{x}{2}-1\Big)+2i \sin \dfrac{x}{2} \cos \dfrac{x}{2} } \\ &= \frac{1}{2 \cos ^2 \dfrac{x}{2}+2i\sin \dfrac{x}{2} \cos \dfrac{x}{2} } \\ &= \frac{1}{2 \cos \dfrac{x}{2}} \cdot \frac{1}{\cos \dfrac{x}{2}+i \sin \dfrac{x}{2}} \\ &=\ldots \end{align}$$

Answer 4

I think your formula is missing a factor of $\frac{1}{2}$.

As you say: the first step is to "realise" the denominator. We have, assuming $x$ is real $$\frac{1}{1+z} = \frac{1}{1+z} \times \frac{1+\overline{z}}{1+\overline{z}} = \frac{1+\overline{z}}{1+z+\overline{z}+z\overline{z}}=\frac{1+\overline{z}}{1+2\Re(z)+|z|^2}=\frac{1+\overline{z}}{2(1+\Re(z))}$$

Since $z=\cos x + \mathrm{i}\sin x$ we have $$\frac{1}{1+\cos x + \mathrm{i}\sin x} =\frac{1+\cos x - \mathrm{i}\sin x}{2(1+\cos x)}=\frac{1}{2}-\frac{\mathrm{i}}{2}\left(\frac{\sin x}{1+\cos x}\right) = \frac{1}{2}-\frac{\mathrm{i}}{2}\tan\left(\frac{x}{2}\right)$$ This last step used a well-know half-angle formula.