Complex Numbers

De Moivre's Theorem

$$(r(\cos \theta +i\sin \theta ){)}^n=r^n(\cos n\theta +i\sin n\theta )$$

Complex Roots

For a complex number

$$z=(r(\cos \theta +i\sin \theta ))$$

The $n^{th}$ roots can be found using the formula

$$z^{\frac{1}{n}}=r^{\frac{1}{n}}\left(\cos \frac{\theta +2k\pi }{n}+i\sin \frac{\theta +2k\pi }{n}\right),k=0,1,2,...,n-1$$

Finding Trig Identities

Trig identities can be found by equating complex numbers and using de moivre's theorem. The examples below are shown for n=2 but the process is the same for any n.

Identities for $f(n\theta )$

Using de moivre's theorem to equate $$\cos 2\theta +i\sin 2\theta =(\cos \theta +i\sin \theta {)}^2$$

Expanding $$(\cos \theta +i\sin \theta {)}^2={\cos }^2\theta +2i\sin \theta \cos \theta -{\sin }^2\theta$$

Equating real and imaginary parts $$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$$ $$\sin 2\theta =2\sin \theta \cos \theta$$

Identities for $f^n(\theta )$

• $z=\cos \theta +i\sin \theta$
• $z^n+z^{-n}=2\cos n\theta$
• $z^n-z^{-n}=2i\sin n\theta$

To find the identity for ${\cos }^2\theta$, start with $z+z^{-1}$, and raise to the power of 2

$$z+z^{-1}=2\cos \theta$$ $$(z+z^{-1}{)}^2=(2\cos \theta {)}^2$$ $$z^2+2+z^{-2}=4{\cos }^2\theta$$

Substituting in for the pairs of $z^n+z^{-n}$

$$(z^2+z^{-2})+2=2\cos 2\theta +2=4{\cos }^2\theta$$ $${\cos }^2\theta =\frac{1}{2}\cos 2\theta +\frac{1}{2}$$