Derive

[Idontunderstand]

I need to derive this:

e^(i*pi) = Cos(pi) + i*sin(pi)

Thank you for any help. I am not even sure where I need to start.

 

[soroban]

Hello, Idontunderstand!

Idon'tunderstand this problem . . .

\(\displaystyle \text{I need to derive this: }\:e^{i\pi} \:=\:\cos(\pi) + i\sin(\pi)\)

The statement is true.
We don't "derive" a true statement, like: .\(\displaystyle 5 \:=\:2+3\)


It follows from the formula: .\(\displaystyle e^{i\theta} \:=\:\cos\theta + i\sin\theta\)

Are you asked to derive this formula?

 

[pka]

I completely agree with Soroban’s sentiment in relpy #2, especially for basic undergraduate complex analysis. However, if you know some advanced topics this can be a very useful exercise.
Recall that \(\displaystyle e^t = \sum\limits_{n = 0}^\infty {\frac{{t^n }}{{n!}}} ,\quad \cos (t) = \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n t^{2n} }}{{\left( {2n} \right)!}}} ,\;\& \,\sin (t) = \sum\limits_{n =0}^\infty {\frac{{\left( { - 1} \right)^n t^{2n + 1} }}{{\left( {2n + 1} \right)!}}} \)

If you notice that the \(\displaystyle \cos\) series is even and \(\displaystyle \sin\) series is odd. Thus use the exponential series \(\displaystyle e^{it}\). It will separate into the sum of two series giving \(\displaystyle e^{it}=\cos(t)+i~\sin(t).\)

 

[Idontunderstand]

Hello, Idontunderstand!

Idon'tunderstand this problem . . .


The statement is true.
We don't "derive" a true statement, like: .\(\displaystyle 5 \:=\:2+3\)


It follows from the formula: .\(\displaystyle e^{i\theta} \:=\:\cos\theta + i\sin\theta\)

Are you asked to derive this formula?

The teacher said derive. I believe that he wants us to prove that statement is true.

 

[galactus]

A popular way to prove this is to use the Taylor series for sin(x) and cos(x).

Adding them then gives the series for \(\displaystyle e^{ix}\)

\(\displaystyle \displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{(2k)!}=cos(x)\)

\(\displaystyle \displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}=sin(x)\)


Of course, multiply the sin series by i. Add them together and you should get:

\(\displaystyle e^{ix}=\displaystyle \sum_{k=0}^{\infty}\frac{(ix)^{k}}{k!}\)


Remember, \(\displaystyle i^{k}\) alternates.

\(\displaystyle i^{2}=-1, \;\ i^{3}=-i, \;\ i^{4}=1, \;\ i^{5}=i, \;\ i^{6}=-1\) and so on.