SOLUTION: Please help me solve the following problem Find all the values of z such that {{{exp(2z-1)=1 }}} My attempt at a solution. {{{e^(2(x+iy)-1)=1}}} {{{e^(2x-1) * e^(i2y)=1}}}

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Please help me solve the following problem Find all the values of z such that {{{exp(2z-1)=1 }}} My attempt at a solution. {{{e^(2(x+iy)-1)=1}}} {{{e^(2x-1) * e^(i2y)=1}}}      Log On


   

Question 1115576: Please help me solve the following problem
Find all the values of z such that
exp%282z-1%29=1+
My attempt at a solution.
e%5E%282%28x%2Biy%29-1%29=1
e%5E%282x-1%29+%2A+e%5E%28i2y%29=1
e%5E%282x-1%29+%2A%28cos2y+%2Bi+sin2y%29=+1
e%5E%282x-1%29+%2Acos2y+%2B+e%5E%282x-1%29+%2Asin2y=+1
Pairing the real and imaginary parts gives
e%5E%282x-1%29+%2Acos2y+=+1+ ----------- (1)
and
e%5E%282x-1%29+%2Asin2y+=+0+ ----------- (2)
Then
sin2y=0
+y=+pi++
since Im trying to find a value of of y such that
+sin2y=0+
and
x=+1%2F2++
Therefore, +z=+1%2F2+%2B+i%28pi%29+%2B+i%282pi%29n
but the book gives the answer as +z=+1%2F2+%2B+i%28pi%29n so I know I'm messing up the imaginary part I'm just not sure how .

Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
At the step sin(2y) = 0, you wrote +y=pi+ but it is probably better to write +y+=+%28n%2Api%29%2F2+ as potential solutions. That solves eq (2), but we need to check this solution in eq (1):
In solving eq (1), let n=0: +e%5E%282x-1%29%2Acos%282%2A%280%2Api%2F2%29%29+=+1+ —> +x=1%2F2+
However, eq (1) must be checked for other values of n:
n=1: +cos%282%2A%281%2Api%2F2%29%29+=++cos%28pi%29+=+-1+ thus n=1 is NOT a solution for eq 1
n=2: +cos%282%2A%282%2Api%2F2%29%29+=+cos%282pi%29+=+1+ n=2 is a solution
n=3: +cos%282%2A%283%2Api%2F2%29%29+=+cos%283pi%29+=+-1+ n=3 is NOT a solution
n=4: +cos%282%2A%284%2Api%2F2%29%29+=+cos%284pi%29+=+1+ n=4 is a solution

The pattern is the solution to eq (1) with +y=n%2Api%2F2+ is only valid for even n:
Thus, the overall solution is +z+=+%281%2F2%29+%2B+i%2An%2Api%2F2+ n = 0, 2, 4, 6, ….


Which reduces to the equivalent (letting m = n/2 just to highlight this step):
+z+=+%281%2F2%29+%2B+i%2Am%2Api+ m = 0,1,2,3,….


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