Complex numbers De Moivre's theorem

[iDimension]

Homework Statement

If
$$C = 1+cos\theta+...+cos(n-1)\theta,$$
$$S = sin\theta+...+sin(n-1)\theta,$$prove that
$$C=\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}} cos\frac{(n-1)\theta}{2} \enspace and \enspace S = \frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}sin\frac{(n-1)\theta}{2}$$

Homework Equations

The Attempt at a Solution

$$C+iS = 1+(cos\theta+isin\theta)+...+(cos(n-1)\theta+isin(n-1)\theta)$$
$$=1+e^{i\theta}+...+e^{i(n-1)\theta}$$
$$=1+z+...+z^{n-1}, \enspace where \enspace z=e^{i\theta}$$
$$=\frac{1-z^n}{1-z}, \enspace if \enspace z\neq1$$
$$=\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in\theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{i\theta}{2}(e\frac{-i\theta}{2}-e\frac{i\theta}{2})}$$
$$=e^{i(n-1)\frac{\theta}{2}}\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}$$
$$=(cos(n-1)\frac{\theta}{2}+isin(n-1)\frac{\theta}{2})\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}$$

Is this correct? I'm not sure about the final solution. The final result is that the real parts equal the imaginary parts?

 

[ehild]

Homework Statement

If
$$C = 1+cos\theta+...+cos(n-1)\theta,$$
$$S = sin\theta+...+sin(n-1)\theta,$$prove that
$$C=\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}} cos\frac{(n-1)\theta}{2} \enspace and \enspace S = \frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}sin\frac{(n-1)\theta}{2}$$

Homework Equations

The Attempt at a Solution

$$C+iS = 1+(cos\theta+isin\theta)+...+(cos(n-1)\theta+isin(n-1)\theta)$$
$$=1+e^{i\theta}+...+e^{i(n-1)\theta}$$
$$=1+z+...+z^{n-1}, \enspace where \enspace z=e^{i\theta}$$
$$=\frac{1-z^n}{1-z}, \enspace if \enspace z\neq1$$
$$=\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in\theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{i\theta}{2}(e\frac{-i\theta}{2}-e\frac{i\theta}{2})}$$
$$=e^{i(n-1)\frac{\theta}{2}}\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}$$
$$=(cos(n-1)\frac{\theta}{2}+isin(n-1)\frac{\theta}{2})\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}$$

Is this correct? I'm not sure about the final solution. The final result is that the real parts equal the imaginary parts?

Yes, your derivation is correct, but the the real part is not equal to the imaginary part.

 

[FactChecker]

Correct. But I have an important word-smithing note:
Throughout the entire post, you should put parentheses around all the angles that you take sins and cos of.
Not
$$=cos(n-1)\frac{\theta}{2}$$
but
$$=cos((n-1)\frac{\theta}{2})$$

 

[Ray Vickson]

Correct. But I have an important word-smithing note:
Throughout the entire post, you should put parentheses around all the angles that you take sins and cos of.
Not
$$=cos(n-1)\frac{\theta}{2}$$
but
$$=cos((n-1)\frac{\theta}{2})$$

Also: do not write ##sin \theta## and ##cos \theta## (ugly and hard to read); instead, write ##\sin \theta## and ##\cos \theta## (easy to read and looks good). To do it, just put a "\" before your sin or cos, so say "\sin" instead of "sin", etc. This holds as well for all the other trig functions, as well as 'log', 'ln', 'lim', 'max', 'min', 'exp' and several others.