Need help on last few steps of complex numbers

Question

If $z=\cos\theta+i\sin\theta$, express $\displaystyle\frac{1}{1+z}$ in the form of $a+bi$.

My working:

$\begin{align}\displaystyle\frac{1}{1+z}&= \frac{1}{1+z}\big(\frac{1-z}{1-z}\big)\\&= \frac{1-z}{1-z^2}\\&=\frac{1-\cos\theta-i\sin\theta}{1-(\cos\theta-i\sin\theta)^2}\\&=\frac{1-\cos\theta-i\sin\theta}{1-(\cos^2\theta-2i\cos\theta\sin\theta+i^2\sin^2\theta)}\\&=\frac{1-\cos\theta-i\sin\theta}{1-\cos^2\theta+2i\cos\theta\sin\theta+\sin^2\theta}\\&=\frac{1-\cos\theta-i\sin\theta}{1-\cos^2\theta+2i\cos\theta\sin\theta+1-\cos^\theta}\\&=\frac{1-\cos\theta-i\sin\theta}{2-2\cos^2\theta+2i\cos\theta\sin\theta}\end{align}$

I'm not sure how to proceed by separating the fraction in order to get my answer in the form $a+bi$.

Any help is appreciated.

Answer 1

$$\frac1{1+z}\cdot\frac{1+\overline z}{1+\overline z}=\frac{1+\overline z}{|1+z|^2}=$$

$$\frac{(1+\cos\theta)-i\sin\theta}{(1+\cos\theta)^2+\sin^2\theta}=\frac{1+\cos\theta}{2(1+\cos\theta)}-\frac{\sin\theta}{2(1+\cos\theta)}i=\frac12-\frac12\frac{\sin\theta}{1+\cos\theta}i$$

Answer 2

hint

$$z=\cos(\theta)+i\sin(\theta) \; \implies$$

$$1+z=2\cos^2(\frac{\theta}{2})+2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$$

$$=2\cos(\frac{\theta}{2})\Bigl(\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})\Bigr)$$

$$=2\cos(\frac{\theta}{2})e^{i\frac{\theta}{2}}$$

Answer 3

As with all complex divisions, you can multiply by the conjugate of the denominator,

$$\frac{1+\bar z}{(1+z)(1+\bar z)}=\frac{1+\Re(z)}{1+2\Re(z)+|z|^2}-i\frac{\Im(z)}{1+2\Re(z)+|z|^2} \\=\frac{1+\cos\theta}{(1+\cos\theta)^2+\sin^2\theta}-i\frac{\sin\theta}{(1+\cos\theta)^2+\sin^2\theta}.$$

As stated, the question does not request to simplify further. (Though this is $\frac12-\frac i2\tan\frac\theta2$.)

Answer 4

why no one likes using picture.. You can see that $1+z=2\cos{\left(\frac{\theta}{2}\right)}\angle\frac{\theta}{2}$. Therefore $\frac{1}{1+z}=\frac{1}{2\cos{\left(\frac{\theta}{2}\right)}}\angle -\frac{\theta}{2}$ a.k.a $\frac{1}{1+z}=\frac{1}{2}-i\frac{1}{2}\tan{\left(\frac{\theta}{2}\right)}$

Answer 5

$$\dfrac1{1+e^{i\theta}}=\frac{e^{-i\theta/2}}{e^{i\theta/2}+e^{-i\theta/2}}=\frac12\frac{\cos\frac\theta2-i\sin\frac\theta2}{\cos\frac\theta2}.$$