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I need to find all solutions to the complex equation $e^{1/z} = \sqrt{e}$ Then I need to show that all these solutions are on the circle $|z-1|=1$ Using the fact that $e^{2\pi i}=1$, I solved the equation to find $z = \frac{2}{1-4ik\pi}$ but that is not what's in the back of my book. Any help from the math community is welcome.

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...and what's what is in your book's back, if may we ask? – DonAntonio Jun 4 '13 at 16:04
z=(2-8kpi*i)/(1+16k²pi²), but when I put my solution set in the equation of the circle, and calculate the absolute value, it comes out to be 1!!! – imranfat Jun 4 '13 at 16:07
I've no idea how they did reach that answer ... but may depend on some specific branch of the square root function or something like that...? Are you sure you copied correctly the equality? – DonAntonio Jun 4 '13 at 16:12
Absolutely, I triple checked. The book's answer made no sense to me, but along it says that"it is easy to show that it lies on the specified circle" so I did not look up at the wrong answer section either. I am pretty confident with this material, so when I found such a different answer in the back→hence this post. – imranfat Jun 4 '13 at 16:17
The book's answer and your answer agree. Just multiply the numerator and the denominator of your answer by $1+4ik\pi$ and then replace $k$ with $-k$ to get the book's answer. – Thomas Andrews Jun 4 '13 at 16:22
up vote 1 down vote accepted

Your answer and the book's answer are essentially the same since: $$\frac{2}{1-4ik\pi} \cdot \frac{1+4ik\pi}{1+4ik\pi}= \frac{2+8ik\pi}{1+16k^2\pi^2}$$

That means your answer for $k$ gives the book's answer for $-k$.

Andre's answer shows why all the solutions are are the circle $|z-1|=1$.

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$$e^{\frac1z}=e^{\frac12}\iff \frac1z-\frac12=2k\pi i\;,\;\;k\in\Bbb Z\;\ldots\ldots$$

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Yes, although your approach slightly different (but cool nonetheless), it comes down to the same answer as what I had. I think something in the back of the book ain't right – imranfat Jun 4 '13 at 16:14
Two things: (1) what book is that, and (2) Be sure to be checking the answer to the correct section in the correwc t chapter and in the correct section. – DonAntonio Jun 4 '13 at 16:16
It is a course book written by the faculty of my school (abroad) so it is not published or anything. Mistakes can happen so to say. Looking at your answer as well as Andre's, I think the back of the book is wrong. – imranfat Jun 4 '13 at 16:33

My calculation gives $z=\frac{2}{1+4k\pi i}$. Of course that is essentially the same as your answer. When we subtract $1$, we get $\frac{1-4k\pi i}{1+4k \pi i}$, which has norm $1$.

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Your answer's exactly the same as mine, but why did you substract one? To check whether the solutions lay on some circle? – DonAntonio Jun 4 '13 at 16:19
Yes, it is the verification that $|z-1|=1$. – André Nicolas Jun 4 '13 at 16:26
Allright, thanks! I am think I am good now. This question is solved... – imranfat Jun 4 '13 at 16:34

to see clearly why the solution set lies on the given circle, first use the transform $w=\frac1z$, the equation becomes $w-\frac12 = 2k\pi i$ thus all solutions for $w$ lie on the "vertical" straight line $\mathfrak{Re}(w)=\frac12$.

under the reverse transform $z=\frac1w$ this line is then mapped to the circle $|z-1|=1$ because $\infty \rightarrow 0$ and the two intersections of the straight line with the unit circle, $\frac12(1 \pm \sqrt{3}i)$ are merely interchanged. it is then easy to show that the three points $0, \frac12(1 \pm \sqrt{3}i)$ all lie at unit distance from 1.

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