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Is there a $z$ for which $z$, $1+i$, $(1+i)z$ and $e^z$ are collinear? There is a close call around $z = .18 + 1.09i$ but I'd like to see a mathematical solution.

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For real $t$, let $z=(1+i)/(1+it)$. I got this by solving $1+i-z = tiz$, that is by requiring that the first three of your numbers are collinear. Now we want $e^z-z$ also a real multiple of $iz$, say. Plotting the imaginary part of $(e^z-z)/(iz)$ we see there is a solution around $t=-0.723$, where $z=0.17457+1.1278 i$. To see that there is such a $t$, note that when $t=0$ this imaginary part is

$ \frac{-\operatorname{e} ^{1} \operatorname{sin} (1)}{2} + 1 - \frac{\operatorname{e} ^{1} \operatorname{cos} (1)}{2} \approx -0.88 < 0 $

and when $t=-1$ it is

$ -\operatorname{sin} (1) + 1 \approx 0.16 > 0 $

so it is zero somewhere between.