# Complex numbers De Moivre's theorem

• iDimension
In summary, the homework equations state that if $\frac{1-z^n}{1-z}e^{i(n-1)\frac{\theta}{2}}$, then $C=\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}} cos\frac{(n-1)\theta}{2}$ and $S=\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}sin\frac{(n-1)\theta}{2}$.
iDimension

## Homework Statement

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

## 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?

iDimension said:

## Homework Statement

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

## 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.

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
$$=cos((n-1)\frac{\theta}{2})$$

FactChecker said:
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.

## What are complex numbers?

Complex numbers are numbers that have both a real and imaginary component. They are expressed in the form a + bi, where a is the real part and bi is the imaginary part, with i being the imaginary unit equal to the square root of -1.

## What is De Moivre's theorem?

De Moivre's theorem states that for any complex number z = r(cos θ + i sin θ), where r is the magnitude and θ is the argument, the nth power of z can be found by raising the magnitude to the nth power and multiplying the argument by n. In other words, (r(cos θ + i sin θ))^n = r^n(cos nθ + i sin nθ).

## How is De Moivre's theorem used in mathematics?

De Moivre's theorem is used in many different areas of mathematics, including complex analysis, trigonometry, and even number theory. It allows for simpler calculations when working with complex numbers, as it provides a way to easily find powers of complex numbers.

## Can De Moivre's theorem be proved?

Yes, De Moivre's theorem can be proved using mathematical induction and basic trigonometric identities. However, it is often accepted as a fundamental theorem in complex analysis without needing a formal proof.

## Are there any practical applications of De Moivre's theorem?

De Moivre's theorem has practical applications in engineering, physics, and other fields where complex numbers are used. It can be used to solve differential equations, model oscillations, and analyze electrical circuits. It also has applications in probability and statistics, particularly in the study of normal distributions.

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