# Help complex equation cos(z) = -isin(z)

1. Jan 9, 2009

### Carl140

1. The problem statement, all variables and given/known data

Let z be a complex number. I want to solve cos(z)= -i*sin(z).

3. The attempt at a solution

Here's my work:

cos(z) = -i*sin(z) implies cos(z) + isin(z) = 0.
Therefore exp(i*z) = 0. Now put z= x+iy then i*z = i*(x+iy) = ix - y, hence
exp(i*z) = exp(ix-y) = exp(ix)*exp(y) =0 but exp(y) is always nonzero so this implies
exp(i*x) = 0 hence cos(x)+i*sin(x) =0 thus cos(x)=0 and sin(x)=0 which is impossible.

So I think there are no solutions. Is this correct?

2. Jan 9, 2009

### chaoseverlasting

No. Here, 0 is also a complex number (0+i0). Now you have cosz + isinz=0+i0. As the real and imaginary parts must be equal, you have two conditions, cosz=0 and sinz=0. That gives you values for z and the total solution is the union of those two solutions.

3. Jan 9, 2009

### D H

Staff Emeritus
No. First off, you are implicitly assuming that cos z and sin z are real. Secondly, the total solution will be the intersection, not union.

Carl140: You're analysis is correct.

4. Jan 10, 2009

### chaoseverlasting

You're right. My mistake. I messed up.