# Is there a proof for the quotient of two values in complex number trigonometry?

Let there be two complex numbers in a complex number plane, one of which is $Z_1=r_1(\cos(\theta_1)+i\cdot \sin(\theta_1))$ and the other $Z_2=r_2(\cos(\theta_2)+i\cdot \sin(\theta_2))$

The formula for the quotient of these two values being

$$\frac{Z_1}{Z_2}=\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i\cdot \sin(\theta_1-\theta_2)]$$

is there a proof for this formula? And how does the division of complex numbers affect this formula?

$$\frac{Z_1}{Z_2}=\frac{r_1e^{i \theta_1}}{r_2e^{i \theta_2}}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}=\frac{r_1}{r_2}\left(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2)\right).$$

• Of course this works as well. The purpose of the proof I provided though was more aimed at proving the theorem by using complex number quotient properties. – joshuaheckroodt Aug 10 '17 at 1:50

The division of complex numbers is more complicated than real numbers, given that for the complex number $\displaystyle \frac{Z_1}{Z_2}$, the conjugate of $Z_2$ is always utilised to calculate the result.

Hence, the proof for the above formula is given by the following:

1. $\displaystyle\frac{Z_1}{Z_2}=\frac{r_1(\cos(\theta_1)+i\cdot \sin(\theta_1))}{r_2(\cos(\theta_2)+i\cdot \sin(\theta_2))}$

from here, it's necessary to multiply the numerator and denominator by the conjugate of $\cos(\theta_2)+i\cdot \sin(\theta_2)$, which is $\cos(\theta_2)-i\cdot \sin(\theta_2)$

1. $\displaystyle \frac{Z_1}{Z_2}=\frac{r_1}{r_2}\cdot \frac{\cos(\theta_1)+i\cdot \sin(\theta_1)}{\cos(\theta_2)+i\cdot \sin(\theta_2)}\cdot\frac{\cos(\theta_2)-i\cdot \sin(\theta_2)}{\cos(\theta_2)-i\cdot \sin(\theta_2)}$

now expand as usual

1. $\displaystyle \frac{Z_1}{Z_2}=\frac{r_1}{r_2}\cdot\frac{\cos(\theta_1)\cos(\theta_2)-\cos(\theta_1)i\cdot \sin(\theta_2)+i\cdot \sin(\theta_1)\cos(\theta_2)-i^2 \sin(\theta_1)\sin(\theta_2)}{\cos^2(\theta_2)-i^2\sin^2(\theta_2)}$

in this particular step, it is important to mention that i is a value with a property such that $i^2=-1$, hence wherever $-i^2$ is present, in the next step those values will become +1. In addition, some knowledge of trigonometric identities is assumed, since substitutions using these identities will be made in the following steps.

1. $\displaystyle \frac{Z_1}{Z_2}=\frac{r_1}{r_2}\cdot \cos(\theta_1)\cos(\theta_2)+\sin(\theta_1)\sin(\theta_2)+i\cdot \sin(\theta_1)\cos(\theta_2)-\cos(\theta_1)i\cdot \sin(\theta_2)$

and finally

1. $\displaystyle \frac{Z_1}{Z_2} = \frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i\cdot \sin(\theta_1-\theta_2)]$

ultimately, the division of complex numbers is the main reason that this proof is done the way it's done. It's nothing special, but it's always cool. Any comments or improvements, feel free to respond.