Here is an interesting problem I'm having some difficulty with:
Find all the complexes z that satisfy the equation:
cos(z) = cosh(z).
The answer should be z=k(pi)(1+i) & z=k(pi)(1-i), but I don't know how to show this.
Any help, folks?
Thanks.
Here is an interesting problem I'm having some difficulty with:
Find all the complexes z that satisfy the equation:
cos(z) = cosh(z).
The answer should be z=k(pi)(1+i) & z=k(pi)(1-i), but I don't know how to show this.
Any help, folks?
Thanks.
consider the function $\displaystyle \cos z = \frac{e^{iz} + e^{-iz} }{2}$ adding $\displaystyle 2 \pi$ to z will make no difference as $\displaystyle \cos (z +2 \pi) = \frac{e^{i(z +2 \pi)} + e^{-i(z +2 \pi)} }{2} \ = \ \frac{e^{iz}e^{i 2\pi} + e^{-iz}e^{-i 2\pi} }{2}$ however $\displaystyle e^{i 2\pi} = 1$ so $\displaystyle \cos z = \cos (z + 2 \pi)$.
Bobak