# Solution of an equation: unique in unit disk

I'm stuck with an exercise. Let $t$ be real, $t > 1$, then:

$te^{t-z}=1$ has a unique solution $z_0$ in the unit disk. Also, $z_0$ is real.

My solution is $z_0=\log(t)+t$ obviously, but if $t > 1$, then this is real, and strictly greater than $1$, hence not in the unit disk? So I do not get behind the existence.

(Logarithm has period $2\pi i$ in the complex plane, and only one of the points $z_0, z_0+2\pi i, z_0-2 \pi i, ...$ can be in the unit circle, since its diameter is $2$, so the uniqueness is no problem.)

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I think what you've written is correct. It looks like something is wrong with the statement of the problem. –  Gerry Myerson Nov 25 '11 at 22:19
Perhaps it was supposed to be $ze^{t−z}=1$? The corrected question might turn out to be similar to the following: math.stackexchange.com/questions/25239/complex-zeros-of-z2ez-z –  Jonas Meyer Nov 26 '11 at 1:44