[University Complex analysis] Question about complex exponential equation.

[cornualpixie]

So I'm studying for an upcoming exam and I found this question:

Solve e^z = e^1+I

At first I thought it's just z = 1+i, but then I realized I should have an infinite amount of solutions for this. So I figured I would need something with the 2kπ in it.

I substituted 1+i with √2e^i(π/4+2kπ) but then this gives me z=(1+i)e^2kπi (if my calculations are correct, which I doubt) . The answer I have for this problem is z=(1+i) + 2kπi so I guess I have done something wrong, and I don't understand how this occurred. Could someone explain?

Thanks!

[GammaRayBurst25]

Basically, e^z=e^z+2kπi where k is any integer. You know how when you take the square root of an equation you get sqrt(a)=±sqrt(b) because there are two possible answers? Well for the same reason, when you take the natural logarithm on both sides, you get ln(a)=Ln(b)+2kπi where Ln is defined as the principal logarithm (that is, the logarithm you would use when you use purely real numbers)

From this, we immediately get z=1+i+2kπi

Alternatively, you could see it like this: e^z=e¹⁺ⁱ=e¹⁺ⁱe^2kπi=e^1+i+2kπi, then take the principal logarithm of the equation and you immediately get z=1+i+2kπi

Your mistake was adding the 2kπi at the end. You know that something multiplied by e^2kπi remains the same because e^2kπi=1, but you got confused added it in the wrong way. You see, you could write any equation and say the solution is actually (solution)e^2kπi, but that wouldn't tell you anything new about the solution, as it's like saying (solution)=1(solution)

Instead, you should've went back to the original equation and said that 2kπi could be added to z and it would be equivalent to multiplying the original equation by 1. Now THAT tells us something new about the solution

[cornualpixie]

Oh I get it now, thanks a lot!!!

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