Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

## Mar 1 De Moivre's Theorem

MathJax TeX Test Page Given a complex number $a + bi$, you can graph it in Cartesian coordinates as (a,b).  So, you can show that the magnitude (or absolute value) $(a^2 + b^2)$ of the products is equal to the product of the magnitudes. In other words, the distance from the center to $(a + bi)$ * the distance from the center to $(c + di)$ = the distance to $(a + bi)(c + di)$.  A complex number can also be expressed in polar coordinates, because $a = r\cos(\theta)$ and $b = i * r\sin(\theta)$, so the angle is $\arctan(\frac{b}{a})$. By multiplying two complex numbers you can see the angle is added. So, $a + bi$ becomes $r(\cos(\theta) + i\sin(\theta))$, which is written as $rcis(\theta)$.

De Moivre's Theorem: $$\dfrac{1}{rcis(\theta)} = \frac{1}{r}cis(-\theta)$$ $$(rcis(\theta))^n = r^n cis(n\theta)$$ $$r_1cis(\theta_1) \cdot r_2cis(\theta_2) = r_1r_2cis(\theta_1 + \theta_2)$$ $$\dfrac{r_1cis(\theta_1)}{r_2cis(\theta_2)} = \dfrac{r_1}{r_2}\cdot cis(\theta_1 - \theta_2)$$ To find the nth root of a complex number, you do a similar thing to the power rule:  $$r^\frac{1}{n} * cis(\dfrac{\theta + 360k}{n}), \text{ with k being all integers from } [0, n-1]$$ Another way to verify this is using this equality. $$e^{i\theta} = cis(\theta)$$ $$\dfrac{1}{rcis(\theta)} = \frac{1}{re^{i\theta}} = \frac{1}{r}e^{-i\theta} = \boxed{\frac{1}{r}cis(-\theta)}$$ $$(r cis(\theta))^n = \left(re^{i\theta}\right)^n = r^ne^{in\theta} = \boxed{r^ncis(n\theta)}$$ $$r_1cis(\theta_1) \cdot r_2cis(\theta_2) = r_1e^{i\theta_1}\cdot r_2e^{i\theta_2} = r_1r_2e^{i\left(\theta_1 + \theta_2\right)} = \boxed{r_1r_2cis(\theta_1 + \theta_2)}$$ $$\dfrac{r_1cis(\theta_1)}{r_2cis(\theta_2)} = \dfrac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}} = \dfrac{r_1}{r_2}e^{i\left(\theta_1 - \theta_2\right)} = \boxed{\dfrac{r_1}{r_2}\cdot cis(\theta_1 - \theta_2)}$$