Saturday, 23 February 2019

Euler's formula for complex z



Given Euler's formula eix=cosx+isinx for xR, how does one extend this definition for complex z? I.e for zC, show that eiz=cos(z)+isin(z)?




I can reexpress z=x+iy for x,yR and then substitute into the above to get eixey=(cosx+isinx)ey, 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)ey+isin(x)ey=(cos(x)+isin(x))ey=eixe+i2y=ei(x+iy)=eiz



where we have used the fact that cosh(y)sinh(y)=ey


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...