# Math Help - Question involving complex numbers

1. ## Question involving complex numbers

Hi I really do not know how to do this.

Write sin2a-cos2ai in modulus argument form
I was thinking of changing sin2a to cos(pi/2 - 2a)..

Will really appreciate the help! Thanks,
J.

2. ## Re: Question involving complex numbers

Hey Tutu.

What is modulus argument form? Is that in terms of re^(i*theta) where r = 1?

3. ## Re: Question involving complex numbers

Originally Posted by Tutu
Hi I really do not know how to do this.

Write sin2a-cos2ai in modulus argument form
I was thinking of changing sin2a to cos(pi/2 - 2a)..

Will really appreciate the help! Thanks,
J.
Notice this is in the Cartesian Form \displaystyle \begin{align*} x + y\,i \end{align*} with \displaystyle \begin{align*} x = \sin{(2a)} \end{align*} and \displaystyle \begin{align*} y = \cos{(2a)} \end{align*}. You need to write it in the modulus-argument form \displaystyle \begin{align*} r\, e^{i\,\theta} \end{align*} with \displaystyle \begin{align*} r = \sqrt{x^2 + y^2} \end{align*} and \displaystyle \begin{align*} \theta = \arctan{\left( \frac{y}{x} \right)} \end{align*}.

4. ## Re: Question involving complex numbers

From what I've learnt, modulus-argument form is the polar form where it is rcis(thetha) where r is positive.
re^i(thetha) is exponential form for me.

The answer of this question is in cis form, not exponential form.

5. ## Re: Question involving complex numbers

Originally Posted by Tutu
From what I've learnt, modulus-argument form is the polar form where it is rcis(thetha) where r is positive.
re^i(thetha) is exponential form for me.

The answer of this question is in cis form, not exponential form.
The calculation of \displaystyle \begin{align*} r \end{align*} and \displaystyle \begin{align*} \theta \end{align*} is identical.

6. ## Re: Question involving complex numbers

Sorry, I do not understand..
Can you show me how you would do it?

7. ## Re: Question involving complex numbers

You already have x and y. Evaluate \displaystyle \begin{align*} r = \sqrt{x^2 + y^2} \end{align*} and \displaystyle \begin{align*} \theta = \arctan{\left( \frac{y}{x} \right)} \end{align*}.