# complex numbers eulers formula

• May 2nd 2012, 05:43 AM
qwerty31
complex numbers eulers formula
Using eulers formula show that sin3x = 3sinx - 4sin^3x
How do I do that?
• May 2nd 2012, 05:53 AM
Prove It
Re: complex numbers eulers formula
No need for Euler's formula.

You know that \displaystyle \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \sum_{r = 0}^n {n\choose{r}}\cos^{n-r}{\theta}\sin^r{\theta} \end{align*} by the Binomial Theorem, and \displaystyle \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \cos{n\theta} + i\sin{n\theta} \end{align*} by DeMoivre's Theorem.

Try applying each of these to \displaystyle \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^3 \end{align*}, setting them equal, and equating real and imaginary parts :)
• May 2nd 2012, 05:58 AM
qwerty31
Re: complex numbers eulers formula
Quote:

Originally Posted by Prove It
No need for Euler's formula.

You know that \displaystyle \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \sum_{r = 0}^n {n\choose{r}}\cos^{n-r}{\theta}\sin^r{\theta} \end{align*} by the Binomial Theorem, and \displaystyle \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \cos{n\theta} + i\sin{n\theta} \end{align*} by DeMoivre's Theorem.

Try applying each of these to \displaystyle \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^3 \end{align*}, setting them equal, and equating real and imaginary parts :)

Thats what I thought (personally, I prefer De Moivres method) but the question specifically stated to use the identites of cos and sin from eulers formula
• May 2nd 2012, 06:12 AM
Prove It
Re: complex numbers eulers formula
Hmm, well let's see...

\displaystyle \displaystyle \begin{align*} e^{i\theta} &= \cos{\theta} + i\sin{\theta} \\ \\ e^{-i\theta} &= \cos{\theta} - i\sin{\theta} \\ \\ e^{i\theta} - e^{-i\theta} &= 2i\sin{\theta} \\ \sin{\theta} &= \frac{e^{i\theta} - e^{-i\theta}}{2i} \\ \\ \sin{3x} &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ \\ 3\sin{x} - 4\sin^3{x} &= 3\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right) - 4\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right)^3 \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} - 4\left(\frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x} }{-8i}\right) \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} + \frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x}}{2i} \\ &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ &= \sin{3x} \end{align*}

Q.E.D.
• May 2nd 2012, 06:14 AM
qwerty31
Re: complex numbers eulers formula
Quote:

Originally Posted by Prove It
Hmm, well let's see...

\displaystyle \displaystyle \begin{align*} e^{i\theta} &= \cos{\theta} + i\sin{\theta} \\ \\ e^{-i\theta} &= \cos{\theta} - i\sin{\theta} \\ \\ e^{i\theta} - e^{-i\theta} &= 2i\sin{\theta} \\ \sin{\theta} &= \frac{e^{i\theta} - e^{-i\theta}}{2i} \\ \\ \sin{3x} &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ \\ 3\sin{x} - 4\sin^3{x} &= 3\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right) - 4\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right)^3 \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} - 4\left(\frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x} }{-8i}\right) \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} + \frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x}}{2i} \\ &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ &= \sin{3x} \end{align*}

Q.E.D.

Thank you very much :D do you by any chance know how to write ( -2 + i)e^(-pi/3) in the form a + bi?
• May 2nd 2012, 06:15 AM
Prove It
Re: complex numbers eulers formula
Quote:

Originally Posted by qwerty31
Thank you very much :D do you by any chance know how to write ( -2 + i)e^(-pi/3) in the form a + bi?

Yes, since \displaystyle \displaystyle \begin{align*} e^{\frac{-\pi}{3}} \end{align*} is a real number, just multiply each term in the brackets by it.
• May 2nd 2012, 06:19 AM
qwerty31
Re: complex numbers eulers formula
Quote:

Originally Posted by Prove It
Yes, since \displaystyle \displaystyle \begin{align*} e^{\frac{-\pi}{3}} \end{align*} is a real number, just multiply each term in the brackets by it.

Damn, sorry, i meant e^(-pi/3 i)
I was thinking about turning the exponential into the form a + bi, then multiplying it by (-2 + i)?
• May 2nd 2012, 06:20 AM
Prove It
Re: complex numbers eulers formula
Quote:

Originally Posted by qwerty31
Damn, sorry, i meant e^(-pi/3 i)
I was thinking about turning the exponential into the form a + bi, then multiplying it by (-2 + i)?

Yes, that's exactly what you would do.
• May 2nd 2012, 06:27 AM
qwerty31
Re: complex numbers eulers formula
Quote:

Originally Posted by Prove It
Yes, that's exactly what you would do.

thank you so much :) one more quick question, when solving square roots (or cube roots etc) of complex numbers, is there any specific patterns of what values of k to choose after equating the angles? I know for square roots the values of k taken are usually 0 and 1, but for cube roots some people take 0, 1, 2 and others take 0, 1, -1?
• May 2nd 2012, 07:01 AM
Prove It
Re: complex numbers eulers formula
Quote:

Originally Posted by qwerty31
thank you so much :) one more quick question, when solving square roots (or cube roots etc) of complex numbers, is there any specific patterns of what values of k to choose after equating the angles? I know for square roots the values of k taken are usually 0 and 1, but for cube roots some people take 0, 1, 2 and others take 0, 1, -1?

Whichever ones give the arguments as \displaystyle \displaystyle \begin{align*} \theta \in (-\pi, \pi] \end{align*}.