Derive

Idontunderstand

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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.
 
Hello, Idontunderstand!

Idon'tunderstand this problem . . .


I need to derive this: eiπ=cos(π)+isin(π)\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: .5=2+3\displaystyle 5 \:=\:2+3


It follows from the formula: .eiθ=cosθ+isinθ\displaystyle e^{i\theta} \:=\:\cos\theta + i\sin\theta

Are you asked to derive this formula?
 
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 et=n=0tnn!,cos(t)=n=0(1)nt2n(2n)!,  &sin(t)=n=0(1)nt2n+1(2n+1)!\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 cos\displaystyle \cos series is even and sin\displaystyle \sin series is odd. Thus use the exponential series eit\displaystyle e^{it}. It will separate into the sum of two series giving eit=cos(t)+i sin(t).\displaystyle e^{it}=\cos(t)+i~\sin(t).
 
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Hello, Idontunderstand!

Idon'tunderstand this problem . . .



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


It follows from the formula: .eiθ=cosθ+isinθ\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.
 
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 eix\displaystyle e^{ix}

k=0(1)kx2k(2k)!=cos(x)\displaystyle \displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{(2k)!}=cos(x)

k=0(1)kx2k+1(2k+1)!=sin(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:

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


Remember, ik\displaystyle i^{k} alternates.

i2=1,   i3=i,   i4=1,   i5=i,   i6=1\displaystyle i^{2}=-1, \;\ i^{3}=-i, \;\ i^{4}=1, \;\ i^{5}=i, \;\ i^{6}=-1 and so on.
 
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