Find z for ez=1+i√3: Solution Explained

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So, the general solution is ##z = (\ln r + 2 \pi k)i + \theta## where ##k = 0, \pm 1, \pm 2, \pm 3, \cdots##In summary, the solution to finding all values of z such that ez=1+i√3 is to first rewrite the equation as ##1 + i \sqrt{3} = r e^{i \theta}##, where r and θ can be easily determined. Then, the general solution is ##z = (\ln r + 2 \pi k)i + \theta## where k is any integer.
  • #1
alexcc17
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Homework Statement


Find all values of z such that ez=1+i√3

Homework Equations


The Attempt at a Solution


I have no idea how to do this. I was going to start with ez=exeiy and try to figure something out from that, but I'm not seeing anything. I checked the solution shown below, but I'm really confused as to how they went from the equation given to the first step. An explanation would be really helpful.
 

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  • #2
alexcc17 said:

Homework Statement


Find all values of z such that ez=1+i√3


Homework Equations





The Attempt at a Solution


I have no idea how to do this. I was going to start with ez=exeiy and try to figure something out from that, but I'm not seeing anything. I checked the solution shown below, but I'm really confused as to how they went from the equation given to the first step. An explanation would be really helpful.

If you write ##1 + i \sqrt{3} = r e^{i \theta}## (where you can easily figure out ##r## and ##\theta##) then the equation is ##e^{z} \equiv e^x e^{iy} = r e^{i \theta}##.
 

What is the equation being solved?

The equation being solved is ez=1+i√3, where e represents the mathematical constant e and z is the unknown variable.

What is the solution to the equation?

The solution to the equation is z = ln(1+i√3), where ln represents the natural logarithm function.

What is the process for solving this equation?

The process for solving this equation involves taking the natural logarithm of both sides of the equation, using properties of logarithms to simplify the equation, and then solving for the variable z.

Why is the natural logarithm used in this solution?

The natural logarithm is used because it is the inverse function of the exponential function, which is represented by e. By taking the natural logarithm of both sides of the equation, the exponential term can be isolated and solved for the variable z.

Can this solution be applied to other equations with the same form?

Yes, this solution can be applied to other equations with the same form, where e is raised to a power and set equal to a complex number. The process for solving would be the same, taking the natural logarithm of both sides and then simplifying to solve for the variable.

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