conjugate of the sum of a real number and a complex number in polar form

Question

I have an expression which I have a hard time understanding how they found the conjugate of $1 + e^{it}$ in

$\frac{1}{1 + e^{it}}$

In order to have a fraction without a complex denominator.

Answer 1

Since$$(e^{it})^\ast=(\cos t+i\sin t)^\ast=\cos t-i\sin t=e^{-it},$$the conjugate of $1+e^{it}$ is $1+e^{-it}$, so your fraction is$$\frac{1+e^{-it}}{(1+e^{it})(1+e^{-it})}=\frac{1+e^{-it}}{2+e^{it}+e^{i-t}}=\frac{1+e^{-it}}{2(1+\cos t)},$$which you're free to rewrite with more trigonometric identities. In fact$$1+e^{it}=2e^{it/2}\cos\tfrac{t}{2}\implies\tfrac{1}{1+e^{it}}=\tfrac12e^{-it/2}\sec\tfrac{t}{2}=\tfrac{1-i\tan(t/2)}{2}.$$

Answer 2

The key fact here is that multiplying a complex number by its conjugate gives a real number, so it makes sense to set the conjugate of $1+e^{it}$ to be $\frac{a}{1+e^{it}}$ for some (yet to be determined) real number $a$.

To do this, use the fact that their moduli have to be equal, i.e. $$|1+e^{it}| =\left| \frac{a}{1+e^{it}} \right|\implies a= |1+e^{it} | ^2 \\ = (1+\cos t)^2 +\sin^2 t = 2(1+\cos t) = 4\cos^2 \frac t2 $$ So, $$(1+e^{it})^*= \frac{4\cos^2\frac t2}{1+e^{it}} $$

Answer 3

Alternative (very informal) approach. The (complex) conjugate of the complex number $z$ is the number $\overline{z}$ that represents the mirror image of $z$ with respect to the $x$-axis.

Further when you have any number $w \in \mathbb{C}$ and any number $r \in \mathbb{R}$ then $(w + r)$ is simply the horizontal translation of $w$, by $r$ units (i.e. if $r < 0,$ then the translation is to the left).

Also, you have that for any two complex numbers $w_1, w_2$ then $\overline{(w_1 + w_2)} ~=~ \overline{w_1} + \overline{w_2}.$

Finally, you have that for any real number $r,$ the complex conjugate $\overline{r} = r.$

Putting this all together, it is immediate that

$$\overline{1 + e^{it}} = 1 + e^{-it}.$$