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I have solved the following problem but would like to double check that I did it properly.

The problem says: Find an expression for $z$ as a function of $w$ in terms of the complex logarithm, where $w=f(z):=2e^z+e^{2z}$, and use it to find all values of $z$ for which $f(z)=3$.

So I have the following:

We have $w=2e^z+e^{2z}$, so taking complex logarithms on both sides we get $$\log(w)=\log(2e^z+e^{2z})=\log(e^z(2+e^z))=\log(e^z)+\log(2+e^z)=z+\log(2+e^z)$$

And I am kind of stuck here because I don't know how to deal with $\log(2+e^z)$.

Any help would be very much appreciated... I know once I have the expression, all I hav to do is substitute $w=3$ into the equation, but I cannot get to that equation.

Thanks!

s1047857
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1 Answers1

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Hint: Let $y=e^z$ and solve the quadratic equation

$$ w= 2y+y^2 $$

See a related problem.

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Mhenni Benghorbal
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    OK, Thanks!

    So basically, I have $y^2+2y-w=0$, which gives $y=\frac{-2\pm\sqrt{4+4w}}{2}=-1\pm\sqrt{1+w}$

    And since $y=e^z$, this gives $z=\log(-1\pm\sqrt{1+w})$

    So for $w=3$, we have that $z=\log(-1\pm\sqrt{4})=-3$ and $1$.

    Is that correct?

     – s1047857 Aug 21 '14 at 17:23
  • @user133971: Yes it is correct. Excellent job. – Mhenni Benghorbal Aug 21 '14 at 19:36