Question involving complex numbers

**Tutu** · November 16th 2012, 10:11 PM

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.

**chiro** · November 16th 2012, 10:22 PM

Hey Tutu.

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

**Prove It** · November 16th 2012, 10:26 PM

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 $\displaystyle \begin{align*} \theta = \arctan{\left( \frac{y}{x} \right)} \end{align*}$ .

**Tutu** · November 16th 2012, 10:42 PM

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.

**Prove It** · November 16th 2012, 11:14 PM

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.

**Tutu** · November 16th 2012, 11:56 PM

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

**Prove It** · November 17th 2012, 12:14 AM

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

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