# Thread: Two complex number problems!

1. ## Two complex number problems!

Hello!

(First problem) Express $sin5x$ in terms of $a) sinx$ and $b) cosx$

Attempt: According to de Moivre's formula - $(cosx+isinx)^n=cos(nx)+isin(nx)$ so:

de Moivre - $(cosx+isinx)^5=cos(5x)+isin(5x)$ and
binomial - $(cosx+isinx)^5=cos^5(x)+5cos^4(x)isin(x)-10cos^3(x)sin^2(x)-10cos^2(x)isin^3(x)+5cos(x)sin^4(x)+sin^5(x)i$

So,
(1) $sin5x=5cos^4(x)sin(x)-10cos^2(x)sin^3(x)+sin^5(x)$

After some arithmetic arrangements and and Pythagorean trigonometric identity I got this: $sin5x=12sin^5(x)-12sin^3(x)+sin(x)$ and task under a) is completed (I think).

But the problem is at task b), so, I don't know how to write expression (1) in terms of cosx function. Is there any identity? Or maybe some other method.

(Second problem) If $z$ is a complex number, what does this equation represent: $|z|-2Re(z)=4$

Attempt: No attempt. I need a complete solution and explanation.

Thank you!

2. Originally Posted by patzer
Hello!

(First problem) Express $sin5x$ in terms of $a) sinx$ and $b) cosx$

Attempt: According to de Moivre's formula - $(cosx+isinx)^n=cos(nx)+isin(nx)$ so:

de Moivre - $(cosx+isinx)^5=cos(5x)+isin(5x)$ and
binomial - $(cosx+isinx)^5=cos^5(x)+5cos^4(x)isin(x)-10cos^3(x)sin^2(x)-10cos^2(x)isin^3(x)+5cos(x)sin^4(x)+sin^5(x)i$

So,
(1) $sin5x=5cos^4(x)sin(x)-10cos^2(x)sin^3(x)+sin^5(x)$

After some arithmetic arrangements and and Pythagorean trigonometric identity I got this: $sin5x=12sin^5(x)-12sin^3(x)+sin(x)$ and task under a) is completed (I think).

But the problem is at task b), so, I don't know how to write expression (1) in terms of cosx function. Is there any identity? Or maybe some other method.

(Second problem) If $z$ is a complex number, what does this equation represent: $|z|-2Re(z)=4$

Attempt: No attempt. I need a complete solution and explanation.

Thank you!

You'll get a complete hint and, perhaps, some explanation. The rest is on you.

Write $z=x+iy\,,\,\,x,y\in\mathbb{R}$ , so

$|z|-2Re(z)=4\Longleftrightarrow \sqrt{x^2+y^2}-2x=4\Longrightarrow x^2+y^2=4x^2-16x+16\Longrightarrow$

$3\left(x-\frac{8}{3}\right)^2-\frac{64}{3}-y^2=-16$ ...and now remember a little analytic geometry and you're done.

Tonio