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Let $f(z)=e^z$ and $g(z)=\log(z)$. Find the preimage and image of the following sets under $f$ and $g$.

a) $A=\{z\in \mathbb{C} \mid \mathrm{Im} z=0\}$

b) $B=\{z\in \mathbb{C} \mid \mathrm{Re} z= \mathrm{Im} z\}$

c) $C=\{z\in \mathbb{C} \mid \mathrm{Re} z > 0,\ \mathrm{Im} z > 0\}$

So for part a) I thought the image for $f(z)$, $z$ is any positive real number, 2$\pi ik$ where $k \in \mathbb{Z}$ because we are looking at $f(z)$ that are in Quad I. However I think I am coming at this problem wrong. Any help with this problem would be great. I asked my professor how he wanted this problem asked and he said to graph the image and preimage of the following sets and giving a brief explanation as to why.

Note: I have tried asking this question before and it is now closed because it was off topic. All my professor is asking to find the image and preimage (range) for the following sets. I am having difficult understand what he is asking. You do not have to do every part. I just need some assistance, so please do not close this problem. If you have any issue please comment or message me and I'll try to help out as much as I can.

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  • $\begingroup$ what is your definition of $\log z?$ Note that we can't know what this function is until we know its domain, and its definition on that domain. $\endgroup$ – zhw. Feb 16 '18 at 23:43
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The method is just solving equations and drawing the lines/planes. Let's see how.

Let's first look at the (pre)image of $f(z)=\exp(z)$ and $g(z)=\log(z)$ in $A=\{z\in \mathbb{C} \mid \mathrm{Im} z=0\}$. We write $z=x+iy$ and note that $A=\{x\mid x\in\mathbb{R} \}=\mathbb{R}$ is just the real line, because $y=\mathrm{Im} z=0$.

Now we easily see that $f(A)=\{f(z)\mid z\in A\}= \mathbb{R}_{>0}$, so $f$ maps $A$ (the real line) to half a real line.

The arrow represent the function sending elements from the arrow-tail set to the arrow-head set.

enter image description here


Now for the preimage $f^{-1}(A)=\{z\mid f(z)\in A\}$, we use that $f(x+iy)=e^xe^{iy}=e^x(\cos(y)+i\sin(y))=e^x\cos(y)+ie^x\sin(y)$ and thus we want $ie^x\sin(y)=0$, which happens for $y=k\pi$ with $k\in\mathbb{Z}$, so the preimage are the lines $x+ ik\pi$, so $f^{-1}(A)=\{x+ ik\pi\mid x\in\mathbb{R},k\in\mathbb{Z}\}$.

enter image description here


Now using that $g(z)=\log(z)=\ln(|z|)+i \text{arg}(z)$ with the argument in $\mathbb{R}$ (this is actually all logarithmic functions $\log(z,t)$ with their argument in $(t,t+2\pi]$ with $t\in\mathbb{R}$, since you didn't specify which argument we work in), we find (note that $0$ is not in the domain of $g$) that $g(A\backslash\{0\})=\{x+ ik\pi\mid x\in\mathbb{R},k\in\mathbb{Z}\}$, because the argument of the real line is $k\pi$ with $k\in\mathbb{Z}$.

enter image description here

If you meant $g(z)=\text{Log}(z)$ with principal argument in $(-\pi,\pi]$, then only $k=0$ and $k=1$ hold:

enter image description here

For any individual logarithmic function $\log(z,t)$, there will only be two neighbouring lines on the right, namely the ones with $y\in (t,t+2\pi]$.


The same way, we find that $g^{-1}(A)= \mathbb{R}_{>0}$, because we want $\mathrm{Im} g(z)=\text{arg}(z)=0$.

enter image description here


Now try the same for $B$: use that $x=\mathrm{Re}z=\mathrm{Im}z=y$, and remember that $$f(x+iy)=e^x(\cos(y)+i\sin(y))$$ and $$g(z)=\text{Log}(z)=\ln(|z|)+i \text{Arg}(z)$$ (I think it's better to use the principal argument). So $f(B)=\{e^x(\cos(x)+i\sin(x))\mid x\in\mathbb{R}\}$, in which we recognize a spiral. $f^{-1}(B)=\{x+i(\frac{\sqrt{2}}{2}+k)\pi\mid k\in\mathbb{Z}\}$, which is a collection of horizontal lines. This is because we need to find solve $e^x\cos(y)=\mathrm{Re}f(z)=\mathrm{Im}f(z)=e^x\sin(y)$, so $y=\frac{\sqrt{2}}{2}+k\pi$.


You do not have to do every part. I just need some assistance.

I hope that it's clear now how to do this. Good luck with the other ones.

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Let's first solve for $f(z)$. The preimage of $T$ under $f$ is a set $S$ so that $$T=f(S)$$or$$S=f^{-1}(T)$$we know that $f(z)=e^{x+iy}=e^x\cos y+ie^x\sin y$. Therefore for any $p+iq\in S$ we have $$f(p+iq)=e^p\cos q+ie^p\sin q\in T$$therefore we must have three cases

Case 1: $T=A$

here we get $$e^p\sin q=0\to \sin q=0\to q=k\pi\qquad,\qquad k\in\Bbb Z$$therefore $$f^{-1}(A)=\{x+iy|y=k\pi\qquad,\qquad k\in\Bbb Z\}$$

Case 2: $T=B$

here we obtain $$e^p\cos q=e^p\sin q\to \sin q=\cos q\to q=k\pi+\dfrac{\pi}{4}\qquad,\qquad k\in\Bbb Z$$therefore $$f^{-1}(B)=\{x+iy|y=k\pi+\dfrac{\pi}{4}\qquad,\qquad k\in\Bbb Z\}$$

Case 3: $T=C$

here we get $$e^p\sin q,e^p\cos q>0\to 2k\pi<q<2k\pi+\dfrac{\pi}{2}\qquad,\qquad k\in\Bbb Z$$therefore $$f^{-1}(C)=\{x+iy|2k\pi<y<2k\pi+\dfrac{\pi}{2}\qquad,\qquad k\in\Bbb Z\}$$also the image of $T$ under $f$ is a set $S$ such that$$S=f(T)$$therefore using a similar approach we have$$f(A)=\Bbb R^+\\f(B)=\{e^xe^{ix}|x\in\Bbb R\}\\f(C)=\{re^{i\theta}|r>1\}$$also $f$ and $g$ are inverse functions i.e. $f\circ g(z)=g\circ f(z)=z$ therefore$$f(A)=g^{-1}(A)=\Bbb R^+\\f(B)=g^{-1}(B)=\{e^xe^{ix}|x\in\Bbb R\}\\f(C)=g^{-1}(C)=\{re^{i\theta}|r>1\}\\f^{-1}(A)=g(A)=\{x+iy|y=k\pi\}\\f^{-1}(B)=g(B)=\{x+iy|y=k\pi+\dfrac{\pi}{4}\}\\f^{-1}(C)=g(C)=\{x+iy|2k\pi<y<2k\pi+\dfrac{\pi}{2}\}$$

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  • $\begingroup$ Why not? Take $z=-\dfrac{1}{e}$ therefore $z\in A$ and $g(z)=\log(-\dfrac{1}{e})=-1+i\pi$ $\endgroup$ – Mostafa Ayaz Feb 17 '18 at 10:11
  • $\begingroup$ You're right about that one. Sorry for that. But still we can't say $f^{-1}(A)=g(A)$, since $0$ is definitely not in the domain of $g$. $\endgroup$ – The Phenotype Feb 17 '18 at 12:49
  • $\begingroup$ It also depends on the argument of the function $\log$. You should at least state what that is. Being the inverse is only that easy in the real numbers, but not in the complex ones. $\endgroup$ – The Phenotype Feb 17 '18 at 13:09
  • $\begingroup$ I didn't take the logarithm as the main branch. If so, you're right and i'm wrong $\endgroup$ – Mostafa Ayaz Feb 17 '18 at 18:48

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