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Describe the image of $x=a$, $y=b$ under the mapping $f(z)= z + e^z $.

So if $z=x+yi$, then $f(x+yi)=x+yi+ e^{x+yi}$. Then with the change of coordinates we get

\begin{align*} &\Rightarrow u=x+e^x\cos(y) \quad\text{and}\quad\ v=y+e^x\sin(y)\\ &\Rightarrow u=a+e^x\cos(y) \quad\text{and}\quad v=y+e^a\sin(y) \end{align*} and now I want to find $y$ in terms of $v$ (from the last equation) and then replace it in $u=a+e^x\cos(y)$. But I don't know how to do it.

Help me please.

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The image of the line $x=a$ is

$$a+iy+e^{a+iy},$$ or $$u=a+e^a\cos y,\\v=y+e^a\sin y,$$ where $y$ acts as a parameter. You can eliminate it by expressing $y$ as a function of $u$, and

$$v=\pm\arccos\frac{u-a}{e^a}+e^a\sqrt{1-\left(\frac{u-a}{e^a}\right)^2}.$$

For the line $y=b$,

$$x+ib+e^{x+ib}$$ gives

$$u=x+e^x\cos b,\\v=b+e^x\sin b.$$

This time, it is not possible to express $x$ in terms of $u$. Instead we can write

$$u=\log\frac{v-b}{\sin b}+(v-b)\cot b.$$

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The mapping $e^z$ takes every line $x=a$ to $e^ae^{iy}$ which specifies a circle ${\bf C}(0,e^a)$. It follows by $z=a+iy$ which changes location with $a$ and $iy$ transforms image part of every point shifted by $y$ as $y$ increasing.

If $a>0$ then the radius of the circle with $a+iy$ makes greater arcs and make spirals shapes around $-1-i\pi$.

If $a<0$ then the radius of the circles are small and curves will be very like to lines $a+iy$ with small curvatures. As $a\to-\infty$ these curves are almost lines $\sim a+iy$.

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  • $\begingroup$ how did you realize about the spiral shapes around -1-i$ \pi $? I mean I just saw them using Mathematica, but how can I know about them analytically? $\endgroup$ – Aaron Martinez Apr 15 '17 at 21:04
  • $\begingroup$ I forgot $\pm$ in $-1\pm i\pi$. It's simple, a circle with parameter $y$ as it rounds and $y$ adds it, makes spiral shape. Perhaps Mathematica helps better geometrically, but analytically is as I said. $\endgroup$ – Nosrati Apr 16 '17 at 2:40
  • $\begingroup$ ok, it's still kinda hard to see it in my mind,you must have lot of experience in this math things :D $\endgroup$ – Aaron Martinez Apr 16 '17 at 4:40
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I'm not sure what you're trying to do, but I really don't think you can write down explicitly $\;y\;$ as function of $\;v\;$ . Instead:

$$f(a,b):=a+bi+e^{a+bi}=\left(a+e^a\cos b\right)+\left(b+e^a\sin b\right)i$$

If we treat the above as parametric equations in the plane, we get:

$$\begin{cases} &x=a+e^a\cos b\implies e^a=\cfrac{x-a}{\cos b}\\{}\\ &y=b+e^a\sin b\implies e^a=\frac{y-b}{\sin b}\end{cases}\;\;\implies\frac{x-a}{\cos b}=\frac{y-b}{\sin b}$$

and you get a straight line , specifically:

$$y=\tan b\cdot x+b-\tan b\cdot a$$

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  • $\begingroup$ I don't think that this answers the question at all, which is about finding the images of axis-parallel lines. $f(a,b)$ is actually a single point and it is shown here that a linear relation holds between the real and imaginary parts of this point, but this doesn't mean at all that there is a straight line: every single point belongs to many straight lines. $\endgroup$ – Yves Daoust Apr 15 '17 at 20:42
  • $\begingroup$ yes, I think the same, I didn't realize this answer it's actually wrong. $\endgroup$ – Aaron Martinez Apr 15 '17 at 20:44

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