Given Euler's formula eix=cosx+isinx for x∈R, how does one extend this definition for complex z? I.e for z∈C, show that eiz=cos(z)+isin(z)?
I can reexpress z=x+iy for x,y∈R and then substitute into the above to get eixe−y=(cosx+isinx)e−y, but I am unsure of how to progress. I am hoping this can be done without appealing to the complex sin and cos.
Thanks.
Answer
Let z=x+iy, and you want to show that eiz=cos(z)+isin(z) i.e. Euler's formula applies to complex z.
I will prove the formula starting from the right hand side.
I don't think you can get away from complex trigonometry, where the cosine/sine of a purely imaginary term is defined as a hyperbolic function as follows:-
cos(ix)=cosh(x)
sin(ix)=isinh(x)
Using the above definitions in conjunction with the sum of angles identities, we have
cos(x+iy)=cos(x)cos(iy)−sin(x)sin(iy)=cos(x)cosh(y)−isin(x)sinh(y)
sin(x+iy)=sin(x)cos(iy)+cos(x)sin(iy)=sin(x)cosh(y)+icos(x)sinh(y)
So we have
cos(x+iy)+isin(x+iy)=cos(x)cosh(y)−cos(x)sinh(y)+i(sin(x)cosh(y)−sin(x)sinh(y))=cos(x)e−y+isin(x)e−y=(cos(x)+isin(x))e−y=eixe+i2y=ei(x+iy)=eiz
where we have used the fact that cosh(y)−sinh(y)=e−y
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