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I have posted the problem below. I know that this question has been asked before but I have a question about the proof.

I wrote the expression in polar form as $f(z)=e^z$ and to check injectivity I assumed that $f(z_1)=f(z_2)$ which gives $e^{z_1}=e^{z_2}$. Is that enough to conclude that $z_1=z_2$? Or should I rewrite the functions as $e^{z_1}=e^{x_1}\cos{y}+ie^{x_1}\sin{y}$? The first case I doesn't consider the restriction on $y$ and then if that's the case then it seems pointless to have converted to complex form.

Thanks! enter image description here

Bernard
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john fowles
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3 Answers3

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You need to say that $e^{z_1}=e^{z_2}$ if and only if $z_1 = z_2 + 2\pi ik$ for some $k\in\Bbb Z$. Now bring in the restriction that $0<y = \text{Im}(z)<2\pi$.

Ted Shifrin
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  • i've also seen this be done for a similar problem where $e^{2i(\theta{_1}-\theta{_2})}=1 \Longrightarrow 2(\theta{_1}-\theta{_2})=2\pi k$. I would have assumed, in the second case, that it would have equaled $0$. Is there something unique about $e$ that makes the function $e^z$ cyclic? Or is it because there's an $i$ in the exponent, which means we have to consider eulers identity in some form, which is what causes $e^z=1$ to have infinite solutions $z$, if there are no restrictions? – john fowles Jul 03 '18 at 18:37
  • The problem I'm referring to is here https://math.stackexchange.com/questions/851812/fx-y-x2-y2-2xy-is-one-to-one-on-the-set-a-consisting-of-all-x-y-wi That problem asked to prove that $f(x,y)=(x^2-y^2,2xy)$ is injective if $x>0$. The complex coordinate form was $f(z)=z^2$. So for that problem could I have just proved that $(z_1)=(z_2)+2\pi ik$, right after I find the complex representation? – john fowles Jul 03 '18 at 18:46
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You can also prove the statement without resorting to complex numbers: suppose $$ (e^a\cos b,e^a\sin b)=(e^c\cos d,e^c\sin d) $$ Then $$ (e^a\cos b)^2+(e^a\sin b)^2=(e^c\cos d)^2+(e^c\sin d)^2 $$ which implies $e^{2a}=e^{2c}$, so $a=c$. Therefore $$ \begin{cases} \cos b=\cos d \\ \sin b=\sin d \end{cases} $$ so $b=d+2k\pi$, for some integer $k$. Can $k$ be nonzero?

With complex numbers: suppose $e^{z}=e^{w}$; then $e^{z-w}=1$. Thus $z-w=2ki\pi$ for some integer $k$. In particular, the real parts of $z$ and $w$ are equal. Can you finish?

egreg
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  • Your proof not using complex numbers makes sense but the proof using complex coordinates still confuses me. In particular, the part where $e^{z-w}=1 \Longrightarrow z-w= 2ki\pi$. I get that if $e^{z-w}=1 \Longrightarrow z-w=0 \Longrightarrow (a+ib)-(c+id)=0 \Longrightarrow (a-c) + i(b-d)=0$. And since you said that the real parts are equal we have $i(b-d)=0$. I figure I need to introduce $\cos$ and $\sin$ but I'm not sure how. It seems that I'm missing something fundamental about complex numbers – john fowles Jul 03 '18 at 23:29
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    @johnfowles $e^z=1$ doesn't imply $z=0$, but only $z=2ki\pi$, for some integer $k$. – egreg Jul 04 '18 at 07:53
  • I just found a proof for this and now I understand. Thanks again! – john fowles Jul 04 '18 at 10:26
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If, as is usual, we set

$z = x + iy, \tag 1$

then of course

$e^z = e^{x + iy} = e^x e^{iy} = e^x(\cos y + i\sin y), \tag 2$

which may be written as the mapping

$f:\Bbb R^2 \to \Bbb R^2 \tag 3$

defined by

$f(x, y) = (e^x \cos y, e^x \sin y) = e^x(\cos y, \sin y). \tag 4$

Now if

$e^{z_1} = e^{z_2}, \tag 5$

then

$e^{z_1 - z_2} = 1 \Longrightarrow z_1 - z_2 = 2 \pi i; \tag 6$

it is now easier to make further progress if we switch to the coordinate representation (1) of the complex numbers $z_1, z_2$, since we need specific information concerning tge $y_i$; in these $xy$-coordinates, we see that (6) yields

$(x_1 - x_2) + i(y_1 - y_2) = (x_1 + i y_1) - (x_2 + i y_2) = z_1 - z_2 = 2 \pi i$ $\Longrightarrow x_1 - x_2 = 0, \; y_1 - y_2 = 2 n \pi, \; n \in \Bbb Z, \tag 7$

whence

$x_1 = x_2, \tag 8$

and now it is easy to see that, with $0 < y_1, y_2 < 2\pi$, the only $n \in \Bbb Z$ such that (7) binds is $0$, so

$y_1 = y_2 \tag 9$

as well. So we see that in fact $e^z$ is injective.

Of course, we may also work directly from the form (4) and write

$e^{x_1}(\cos y_1, \sin y_1) = e^{x_2}(\cos y_2, \sin y_2 \Longrightarrow e^{2x_1}(\cos^2 y_1 + \sin^2 y_1) = e^{2x_2} (\cos^2 y_2 + \sin^2 y_2 )$ $\Longrightarrow e^{2 x_1} = e^{2 x_2} \Longrightarrow 2x_1 = 2x_2 \Longrightarrow x_1 = x_2; \tag{10}$

then it follows that

$(\cos y_1, \sin y_1) = (\cos y_2, \sin y_2) \Longrightarrow \cos y_1 = \cos y_2, \; \sin y_1 = \sin y_2; \tag{11}$

the only way this may bind with $0 < y_1, y_2 < 2\pi$ is if

$y_1 = y_2, \tag{12}$

thus establishing the injectivity of $f(x, y)$ on the set where $0 < y_1, y_2 < 2\pi$.

Robert Lewis
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