# Thread: Complex number trig question.

1. ## Complex number trig question.

Show that

$\displaystyle \frac{1+sin\theta+icos\theta}{1+sin\theta-icos\theta}=sin\theta+icos\theta$

and hence that

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5$ $\displaystyle +i(1+sin\frac{\pi}{5}-icos\frac{\pi}{5})^5=0$

I have done the first part and am going round in circles trying to show the second part, I was trying to make use of de moivre's theorem but to no avail. Please help.

2. EDIT: Oops, I misread and thought you were saying you were going in circles with the first part. Sorry

3. Like I said, I have done the first part, just wasn't able to do the second, and would like help for that part, thanks.

4. Hello,

Trying to solve the second problem, I made a few manipulations, but I've seemed to arrive at a contradiction.

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+i(1+sin\f rac{\pi}{5}-icos\frac{\pi}{5})^5=0$

Using $\displaystyle i^5=i$

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+(i+isin\f rac{\pi}{5}+cos\frac{\pi}{5})^5=0$

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5})^5$

Using $\displaystyle \sqrt[5]{-1}=-1$

$\displaystyle 1+sin\frac{\pi}{5}+icos\frac{\pi}{5}=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5})$

$\displaystyle 1+sin\frac{\pi}{5}+icos\frac{\pi}{5}+i+isin\frac{\ pi}{5}+cos\frac{\pi}{5}=0$

$\displaystyle 1+sin\frac{\pi}{5}+cos\frac{\pi}{5} +i(1+sin\frac{\pi}{5}+cos\frac{\pi}{5})=0$

which only holds if the real and imaginary parts are equivalently 0. But because $\displaystyle 1+sin\frac{\pi}{5}+cos\frac{\pi}{5}$ is not, it's weird. My calculator says that I'm wrong.

5. Originally Posted by Dfrtbx
Hello,

Trying to solve the second problem, I made a few manipulations, but I've seemed to arrive at a contradiction.

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+i(1+sin\f rac{\pi}{5}-icos\frac{\pi}{5})^5=0$

Using $\displaystyle i^5=i$

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+(i+isin\f rac{\pi}{5}+cos\frac{\pi}{5})^5=0$

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5})^5$

Using $\displaystyle \sqrt[5]{-1}=-1$

$\displaystyle 1+sin\frac{\pi}{5}+icos\frac{\pi}{5}=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5})$ The contradiction comes here, where you are taking the fifth root of both sides of an equation. A complex number has five fifth roots, and you can't assume that you are getting the same one on both sides of the equation. To take a much simpler example, (–1)^2 = 1^2, but you can't take the square root of both sides and conclude that –1 = 1.

$\displaystyle 1+sin\frac{\pi}{5}+icos\frac{\pi}{5}+i+isin\frac{\ pi}{5}+cos\frac{\pi}{5}=0$

$\displaystyle 1+sin\frac{\pi}{5}+cos\frac{\pi}{5} +i(1+sin\frac{\pi}{5}+cos\frac{\pi}{5})=0$

which only holds if the real and imaginary parts are equivalently 0. But because $\displaystyle 1+sin\frac{\pi}{5}+cos\frac{\pi}{5}$ is not, it's weird. My calculator says that I'm wrong.
To solve the second problem, notice that $\displaystyle \sin\theta+i\cos\theta = i(\cos\theta-i\sin\theta) = i(\cos(-\theta) + i\sin(-\theta)).$ So $\displaystyle \frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=i(\cos(-\theta) + i\sin(-\theta)).$ Now put $\displaystyle \theta = \pi/5$, take the fifth power of both sides and use de Moivre's theorem.