Complex Numbers 2

1. Sep 26, 2007

Oblio

Consider the complex number z=e$$_{i\theta}$$ = cos$$\theta$$+isin$$\theta.$$ By evaluating z$$^{2}$$ two different ways, prove the trig identities cos2$$\theta$$ = cos$$^{2}$$$$\theta$$ - sin$$^{2}$$$$\theta$$ and sin2$$\theta$$ = 2sin$$\theta$$cos$$\theta$$.

A question about the approach to this question:
How do you guys approach the task of 'evaluating' something, when told to do so, like here.
I find myself doing random manipulations without knowledge of whether the road I'm on is even close to the correct path or not.
Evaluate seems like such a general instruction...

Anyways,
If I square z I get;
z$$^{2}$$ = cos$$\theta$$$$^{2}$$ + i$$^{2}$$sin$$\theta$$$$^{2}$$

z$$^{2}$$ = cos$$\theta$$$$^{2}$$ - sin$$\theta$$$$^{2}$$

If I sub z$$^{2}$$ back in I get 0 so that's wrong.

Is the use of the 'e definition' necessary for this? I kind of want to see how I ought to be approaching this 'evaluation'.

Thanks, as always

2. Sep 26, 2007

neutrino

How??

If $$z = \cos{\theta}+\imath\sin{\theta}$$, then $$z^2 = \left(\cos{\theta}+\imath\sin{\theta}\right)^2 \neq \cos^2{\theta}+\imath^2\sin^2{\theta}$$.

Square that. Then use the other way of evaluating a complex number raised to some power, i.e., de Moiver's formula. Equate the two.

Last edited: Sep 26, 2007
3. Sep 26, 2007

Oblio

Am I allowed to do this?

isin$$\theta$$$$^{2}$$

=-sin$$\theta$$$$^{2}$$ ?

4. Sep 26, 2007

neutrino

The way you have written is quite ambiguous. But if you mean $$\left(\imath\sin{\theta}\right)^2$$, then it is equal to $$-\sin^2{\theta}$$.

P.S. You need not enclose each term of LaTeX code within 'tex' tags. It is enough if you do it for each line of code. Example: Enclosing this - \left(\imath\sin{\theta}\right)^2 - within tex tags will produce $$\left(\imath\sin{\theta}\right)^2$$.

5. Sep 26, 2007

Oblio

lol thanks!
It just does that when I click the latex icon each time.. your saying you write it all out manually?

6. Sep 26, 2007

Oblio

I'm still stuck with:

z$$^{2}$$ = cos$$^{2}$$$$\theta$$ + 2isin$$^{2}$$$$\theta$$ - sin$$^{2}$$$$\theta$$

You'll see I still did it the long way if you quote:P

7. Sep 26, 2007

neutrino

While I mostly do it manually(click n' latex is only a recent feature), I'm not asking you to do it manually.

When I click on any icon, what I get is that piece of code (say, \frac{}{} for fraction) within tex codes. The cursor is still within the those tags, so when I click on another icon, only its code is printed within the same set of tags. When I want to do a new line, I move the cursor outside the first set of tex tags.

8. Sep 26, 2007

neutrino

Shouldn't it be $$\cos^2{\theta} + 2\imath\cos{\theta}\sin{\theta} - \sin^2{\theta}$$?

9. Sep 26, 2007

Oblio

yeah, sorry. Typo

10. Sep 26, 2007

neutrino

Did you also find the other expression using de Moivre's theorem/formula?

11. Sep 26, 2007

Oblio

oh i equate the 2 just for one of the two im looking for? I thought that was for the other equation

12. Sep 26, 2007

neutrino

What do you get when you equate the two?

13. Sep 26, 2007

Oblio

I've gotten to

cos2$$\theta$$=cos2$$\theta$$-isin2$$\theta$$+2isin$$\thetacos\theta$$

14. Sep 26, 2007

Oblio

im realizing now thats wrong isnt it..

15. Sep 26, 2007

neutrino

When is a + ib = c + id? (a,b,c,d are real)

16. Sep 26, 2007

Oblio

whennn...
a=c, b=d?

17. Sep 26, 2007

neutrino

Right.

So when you equate cos(2t) + isin(2t) and cos^2(t) + 2isin(t)cos(t) - sin^2(t)...?

18. Sep 26, 2007

Oblio

well if i use what im trying to prove,
cos(2t) = cos^2(t)-sin^2(t)
and sin(2t) = 2sintcost

19. Sep 26, 2007

Oblio

but doesnt this method require that which im proving, to prove it?

20. Sep 26, 2007

neutrino

No. You're just equating the real and imaginary parts of the complex numbers. All you're using is the definition of equality of two complex numbers. (a = c, b = d)