How do we go about doing this?
$e^z = e \implies e^{x+iy} = e \implies x+iy =1$
Obviously $(1,0)$ works, but what else?
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I guess nothing else – the_candyman Oct 27 '14 at 08:33
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2Hint: Think about $2\pi i$ – Chris Culter Oct 27 '14 at 08:35
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1The function $e^{z}$ is an $2\pi i$-periodic function: $e^{z}=e^{z+2k\pi i}, ;;k\in\mathbb{Z}.$ – M. Strochyk Oct 27 '14 at 08:36
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1If $e^z=e$ then $e^{z-1}=1$, so $e^w=1$ if $w=z-1$. Can you find all solutions of $e^w=1$? – bof Oct 27 '14 at 08:37
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1Hint: $e^{x+iy}=e^x(\cos y + i\sin y)=e$ – Lehs Oct 27 '14 at 08:38
3 Answers
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Observe that :$e^z = e = e^1 \iff z - 1 = 2n\pi i \iff z = 1 + 2n\pi i, n \in \mathbb{Z}$.
DeepSea
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$ e^z = e $ , z = x +i*y
=> $|e^z| = |e^{x+i*y}| = e^x = e $ => x= 1 (here the exponential is real, you have injectivity)
Now we get : $ e^z = e*e^{iy} = e $ => $e^{iy} = 1$ => $y = 2k\pi$ , k integer
Let: $z_o$ = 1 + $2ik\pi$ , $e^{z_o} = e^{1+2ik\pi} = e $
So all solutions are { 1+ $2ik\pi$ , k integer}
mvggz
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$$e=e^z=e^{x+iy}=e^x(\cos y+i\sin y)$$ Therefore considering modulus of both sides $$e=e^x,$$ and considering imaginary parts $$0=\sin y.$$ Hence $$x=1,\, y=2n\pi,\,\forall n\in\mathbb{N}.$$
Bumblebee
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I don't see how you deduced that $e = e^x$ from $e=e^x(\cos y+i\sin y)$ and why $\sin y = 0$ – logic Oct 27 '14 at 18:54
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$e=e^x\cos y + e^xi\sin y$ so $Re(e^z) = e^x\cos y$, $Im(e^z)=e^x\cos y \implies e=e^x\cos y, 0=e^x\sin y \implies \sin y =0 \implies y = 2\pi k \implies x = 1\cos(2\pi k) = 1$ Thus $x=1,y=2\pi k$. So $z = 1+2\pi k i$ – logic Oct 27 '14 at 19:10
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if we consider the modulus of both sides, we can obtain $|e^z|=|e^x(\cos y+i\sin y|=e^x\sqrt{\cos^2y+\sin^2y}=e^x$ and also equating imaginary parts of both sides gives $\Im e=e^x\sin y=0.$ – Bumblebee Oct 28 '14 at 17:11