how do you find the zeros of cos z

Question : Find the zeros of cos z

Solution :

zeros of cos z are obtained by solving

cos z = 0

Put cos z = (e^(iz) + e^(−iz))/2

e^(iz) + e^(−iz)=0

multiplying e^(iz) both sides

 e(2iz)+1=0

e^(2iz) = −1

put z=x+iy, i^2=-1

e^(2ix −2y)=-1

e^(2ix )=-e^(2y)

We shall compare real and imaginary parts both sides

Cos(2x) = −e^(2y) , sin(2x)  = 0

Solving second equation sin(2x)  = 0 we get 2x = πk, k ∈ Z or x = πk/2, k ∈ Z

Put x = πk/2 into first equation

cos(πk) = −e^(2y)

(−1)^k = −e^(2y) , k ∈ Z

1 = e^(2y) , k=±1, ±3, . .,.

y =0

 Thus  x = π(2n + 1)/2, y =0 , n ∈ Z. That is zeros of cos z are zn=x+iy = π(2n + 1)/2+i0, n ∈ Z