Question 1115576
At the step  sin(2y) = 0,  you wrote {{{ y=pi }}}  but  it is probably better to write  {{{ y = (n*pi)/2 }}} as <em>potential</em> solutions.   That solves eq (2), but we need to check this solution in eq (1):

In solving eq (1),  let n=0:  {{{ e^(2x-1)*cos(2*(0*pi/2)) = 1 }}} —>   {{{ x=1/2 }}}
However, eq (1) must be checked for other values of n:
n=1:   {{{ cos(2*(1*pi/2)) =  cos(pi) = -1 }}}  thus n=1 is NOT a solution for eq 1
n=2:  {{{ cos(2*(2*pi/2)) = cos(2pi) = 1 }}}  n=2 is a solution
n=3:  {{{ cos(2*(3*pi/2)) = cos(3pi) = -1 }}}  n=3 is NOT a solution
n=4:  {{{ cos(2*(4*pi/2)) = cos(4pi) = 1 }}}   n=4 is a solution<br>

The pattern is the solution to eq (1) with {{{ y=n*pi/2 }}} is only valid for  even n:
Thus,  the overall solution is {{{ z = (1/2) + i*n*pi/2 }}}   n = 0, 2, 4, 6, ….   <br>


Which reduces to the equivalent (letting m = n/2 just to highlight this step):
   {{{ z = (1/2) + i*m*pi }}}   m = 0,1,2,3,….  <br>


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