Finding the Solutions to exp(z) = 1 with Complex Numbers

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The equation ez = 1 is solved by expressing z as a complex number, leading to the form ea(cos(b) + isin(b)) = 1. Since ea must be positive, it follows that sin(b) = 0, resulting in b being n2π or π + n2π. The requirement for cos(b) to be positive restricts b to n2π. Consequently, the real part a must equal 0, yielding the solution z = i2πn. The discussion highlights a common oversight in factoring the imaginary unit i with b.
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Homework Statement



solving the equation ez = 1 , where z is complex



The Attempt at a Solution




This is my attempt at the solution

ez = ea+ib

and then i can write it as ea(cos(b)+isin(b))= 1
ea has to be larger than 0 and therefor i*sin(b) = 0 → b can either be n2\pi or \pi + n2\pi
and cos(b) has to be positive because ea is positive, and the only posible b is n2\pi. And because a is the real part a has to be equal to 0. I really can't see where I'm missing something because in the solution z = in2\pi
 
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fraggy said:

Homework Statement



solving the equation ez = 1 , where z is complex



The Attempt at a Solution




This is my attempt at the solution

ez = ea+ib

and then i can write it as ea(cos(b)+isin(b))= 1
ea has to be larger than 0 and therefor i*sin(b) = 0 → b can either be n2\pi or \pi + n2\pi
and cos(b) has to be positive because ea is positive, and the only posible b is n2\pi. And because a is the real part a has to be equal to 0. I really can't see where I'm missing something because in the solution z = in2\pi

All fine. So b has to be 2*pi*n and a has to be 0. So z=a+i*b=i*2*pi*n. What's the problem?
 
Ok i got it now thx. I forgot to factor in the i with b. Wow really simple misstake
 
misstake is a mistake
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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