# Math Help - calculation method??

1. ## calculation method??

Dear all

eq url：
http://i.imgur.com/1SVsrWr.jpg

than k u for ur help.

*************
Mattew

2. ## Re: calculation method??

$\Large x + \imath y = \rho e^{\imath \theta}$

where
$\large \rho = \sqrt{x^2 + y^2}$
$\large \theta = \arg(x,y) = \arctan(y,x)$ where $\arctan(y,x)=\arctan\left(\frac y x\right)$ with the result adjusted for the correct quadrant.

Then $\Large \left(\rho e^{\imath \theta}\right)^k = \rho^k e^{\imath k \theta}$

Also if you have two complex numbers in polar form their product is given by

$\Large \left(\rho_1 e^{\imath \theta_1}\right)\left(\rho_2 e^{\imath \theta_2}\right)=\rho_1 \rho_2 e^{\imath(\theta_1+\theta_2)}$

that should be all you need to do your calculation

3. ## Re: calculation method??

Very clear !!!!
thank u for ur clear answer.

************
Matthew ^^
Originally Posted by romsek
$\Large x + \imath y = \rho e^{\imath \theta}$

where
$\large \rho = \sqrt{x^2 + y^2}$
$\large \theta = \arg(x,y) = \arctan(y,x)$ where $\arctan(y,x)=\arctan\left(\frac y x\right)$ with the result adjusted for the correct quadrant.

Then $\Large \left(\rho e^{\imath \theta}\right)^k = \rho^k e^{\imath k \theta}$

Also if you have two complex numbers in polar form their product is given by

$\Large \left(\rho_1 e^{\imath \theta_1}\right)\left(\rho_2 e^{\imath \theta_2}\right)=\rho_1 \rho_2 e^{\imath(\theta_1+\theta_2)}$

that should be all you need to do your calculation