complex numbers eulers formula

**qwerty31** · May 2nd 2012, 05:43 AM

Using eulers formula show that sin3x = 3sinx - 4sin^3x
How do I do that?

**Prove It** · May 2nd 2012, 05:53 AM

No need for Euler's formula.

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

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

**qwerty31** · May 2nd 2012, 05:58 AM

Originally Posted by Prove It

No need for Euler's formula.

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

Try applying each of these to $\displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^3 \end{align*}$ , 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

**Prove It** · May 2nd 2012, 06:12 AM

Hmm, well let's see...

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

Q.E.D.

**qwerty31** · May 2nd 2012, 06:14 AM

Originally Posted by Prove It

Hmm, well let's see...

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

Q.E.D.

Thank you very much

do you by any chance know how to write ( -2 + i)e^(-pi/3) in the form a + bi?

**Prove It** · May 2nd 2012, 06:15 AM

Originally Posted by qwerty31

Thank you very much

do you by any chance know how to write ( -2 + i)e^(-pi/3) in the form a + bi?

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

**qwerty31** · May 2nd 2012, 06:19 AM

Originally Posted by Prove It

Yes, since $\displaystyle \begin{align*} e^{\frac{-\pi}{3}} \end{align*}$ 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)?

**Prove It** · May 2nd 2012, 06:20 AM

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.

**qwerty31** · May 2nd 2012, 06:27 AM

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?

**Prove It** · May 2nd 2012, 07:01 AM

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 $\displaystyle \begin{align*} \theta \in (-\pi, \pi] \end{align*}$ .

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