Exponential Form of e^z for z = 4e^(i*pi/3)

MaxManus
Messages
268
Reaction score
1

Homework Statement



write e^z in the form a +bi
z = 4e^(i*pi/3)

---------------------------------------
My guess:

z = 4*(cos(pi/3) + i*sin(pi/3))

e^z = e^(4*(cos(pi/3) + i*sin(pi/3))) = e^(4*cos(pi/3))*(cos(4*sin(pi/3)) + i*sin(4*sin(pi/3)))

but the solution says

e^(2)*(cos(2*sqrt(3)) + i*sin(2*sqrt(3)))
 
Physics news on Phys.org
MaxManus said:

Homework Statement



write e^z in the form a +bi
z = 4e^(i*pi/3)

---------------------------------------
My guess:

z = 4*(cos(pi/3) + i*sin(pi/3))

e^z = e^(4*(cos(pi/3) + i*sin(pi/3))) = e^(4*cos(pi/3))*(cos(4*sin(pi/3)) + i*sin(4*sin(pi/3)))

but the solution says

e^(2)*(cos(2*sqrt(3)) + i*sin(2*sqrt(3)))

Well, \sin(\pi/3)=\frac{\sqrt{3}}{2}, and I'm sure you know what \cos(\pi/3) is...:wink:
 
Thanks for the help.
--------------------------------------
z = 4e^(i*pi/3)
z = 4*(cos(pi/3) + i*sin(pi/3))
z = 2 + 2*sqrt(3)
e^z = e^(2)*(cos(2*sqrt(3)) + i*sin(2*sqrt(3))
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top