Power of Complex numbers proof

[libelec]

Homework Statement

Prove that $$\left( {\frac{{1 + \cos x + i\sin x}}{{1 + \cos x - i\sin x}}} \right)^n} = \cos nx + i\sin nx$$

The Attempt at a Solution

I thought it would be a good idea calling z = 1 + cos x + i*sen x, because then 1 + cos x - i*sen x would be $$\bar{z}$$, and then I would have something of the form $${\left( {\frac{z}{{\bar z}}} \right)^n} = {\left( {\frac{{{z^2}}}{{{{\left| z \right|}^2}}}} \right)^n}$$. If I use the Euler form for complex numbers, being $$\varphi$$ the argument of z, then z$$\left| z \right|{e^{i\phi }}$$.

So $${\left( {\frac{{{z^2}}}{{{{\left| z \right|}^2}}}} \right)^n} = {\left( {\frac{{{{(\left| z \right|{e^{i\phi }})}^2}}}{{{{\left| z \right|}^2}}}} \right)^n} = {\left( {\frac{{{{\left| z \right|}^2}{e^{i2\phi }}}}{{{{\left| z \right|}^2}}}} \right)^n} = {\left( {{e^{i2\phi }}} \right)^n} = {e^{i2n\phi }}\$$. Which would all work just fine with the proof if I could somehow prove that $$\varphi$$ = x/2.

My question is, is this OK? And if it is OK, how can I prove that $$\varphi$$ = x/2?

Thanks

 

[Mark44]

Aren't the numerator and denominator equal to 1 + e^ix and 1 + e^-ix respectively? That's the direction I would go. And the right side is e^i*nx = (e^{i x})ⁿ = (cos x + i sin x)ⁿ,

One comment about your work: You have the exponent wrong on the right side. Your equation should be as follows:
$${\left( {\frac{z}{{\bar z}}} \right)^n} = {\left( {\frac{{{z^2}}}{{{{\left| z \right|}^2}}}} \right)^{n/2}}$$

 

[Hurkyl]

One comment about your work: You have the exponent wrong on the right side. Your equation should be as follows:
$${\left( {\frac{z}{{\bar z}}} \right)^n} = {\left( {\frac{{{z^2}}}{{{{\left| z \right|}^2}}}} \right)^{n/2}}$$

That doesn't look right...

 

[libelec]

Aren't the numerator and denominator equal to 1 + eix and 1 + e-ix respectively? That's the direction I would go. And the right side is ei*nx = (ei x)n = (cos x + i sin x)n,

Yes, that's basically what I did. Only that I recalled the whole numerator z so that I could deal with the division.

One comment about your work: You have the exponent wrong on the right side. Your equation should be as follows:

No, why? $$\frac{z}{{\bar z}} = \frac{{{z^2}}}{{{{\left| z \right|}^2}}}$$

 

[mheslep]

Aren't the numerator and denominator equal to 1 + e^ix and 1 + e^-ix respectively? That's the direction I would go. And the right side is e^i*nx = (e^{i x})ⁿ = (cos x + i sin x)ⁿ,

One comment about your work: You have the exponent wrong on the right side. Your equation should be as follows:
$${\left( {\frac{z}{{\bar z}}} \right)^n} = {\left( {\frac{{{z^2}}}{{{{}^2}}}} \right)^{n/2}}$$

That doesn't look right...

It is not right. Try, e.g., n=2 for z=(cos x + i sin x)ⁿ and we have

(cos x + i sin x)²= $$(cos^2x-sin^2x) + i(2cosxsinx)$$, but
$$\left| z \right|^2 = \left| cos x + isinx \right|^2=1$$

 

[Mark44]

That doesn't look right...

[quote Originally posted by Mark44]
One comment about your work: You have the exponent wrong on the right side. Your equation should be as follows:
$${\left( {\frac{z}{{\bar z}}} \right)^n} = {\left( {\frac{{{z^2}}}{{{{|z|}^2}}}} \right)^{n/2}}$$
[/quote]
I agree. I didn't notice that the conjugate in the denominator on the left had changed to the square of the magnitude in the expression on the right.

 

[Mark44]

Back to the original problem...
$$Prove~that~\left( {\frac{{1 + \cos x + i\sin x}}{{1 + \cos x - i\sin x}}} \right)^n} = \cos nx + i\sin nx$$

$$\left( {\frac{{1 + \cos x + i\sin x}}{{1 + \cos x - i\sin x}}} \right)^n} = \left(\frac{1 + e^{ix}}{1 + e^{-ix}}\right)^n$$

Inside the parentheses, multiply numerator and denominator by e^ix. Then factor e^ix out of the two terms in the numerator and cancel the factor that is common to numerator and denominator. It's then easy to show that what you have left equals cos(nx) + isin(nx) = e^inx = (e^ix)ⁿ.

 

[libelec]

EDIT: I hadn't seen that comment from Mark44.

Thanks, I wouldn't have thought that.

 

[libelec]

Edit.