I need to find and sketch the image of the straight line $z = (1+ia)t +aib$, where $-\infty < t < + \infty$, $a,b\in \mathbb{R}$, and $a \neq 0$, under the map $w = e^{z}$.

In order to accomplish this thus far, I have substituted the expression for $z$ into the expresison for $w$ to yield

$\displaystyle w = e^{z} = e^{(1+ia)t+ib} = e^{(1+ia)t}e^{ib}=e^{t}e^{i(at+b)} = e^{t}(\cos(at+b)+i\sin(at+b))$

What I have to do next is eliminate $t$ in order to put the answer in polar coordinates. So, I let $\varphi = at+b$, and then $t = \frac{\varphi}{a}-\frac{b}{a}$.

Also, I am told that after eliminating $t$, the answer should be $r = ce^{\varphi/a}$, where $c = e^{-b/a}$, which, upon substituting my expression for $t$ into $e^{t}(\cos(at+b)+i\sin(at+b))$ is what I get for $r$: $w = e^{\frac{\varphi}{a}-\frac{b}{a}}(\cos(\varphi)+i\sin(\varphi)) = e^{\frac{\varphi}{a}-\frac{b}{a}}e^{i\varphi}$.

But what about this $e^{i\varphi}$ part? Is this part of the image under the map? Do I have the correct image under the map? And how do I sketch it once I do have the correct image?

I am very, very confused!

  • $\begingroup$ Your approach is not wrong, but generally speaking it isn't a good idea to try to eliminate the variable ($t$) which controls the parametrization of the domain you want to map. Better would be to break it down into cases, according to what $a$ and $b$ might be and then just plot the result under the map $\exp(z)$. It isn't going to be very different from sketching the graph of a function in Calculus. Only in this case you would be plotting against the Complex plane. $\endgroup$ – Yiannis Galidakis Feb 24 '16 at 11:02

You want the answer in the form $r = $ something. That means you need to consider only modulus of $w$. Indeed $$r = \lvert w \rvert = \exp t = \exp\frac{\varphi - b}{a}$$ which is what you want.

The $\exp{i\varphi} = \exp{i(at+b)}$ part indicates that argument of the image rotates with respect to origin, anticlockwise assuming $a > 0$. In that case, $r$ continually increases exponentially, and together with rotation, yields a anticlockwise spiral centered at and away from origin. If $a < 0$, then it's a clockwise spiral centered at and away from origin.

  • $\begingroup$ what I don't understand though is why do I want the answer in the form $r = $ something? As in what in the problem tells me that this is what I'm supposed to end up with? $\endgroup$ – ALannister Feb 24 '16 at 11:01
  • $\begingroup$ @JessyCat, Oh I'm sorry. I thought the question asked for it directly. You don't need the first paragraph of my answer then. Add second paragraph to what you have done so far, and you get the complete answer. $\endgroup$ – Saibal Feb 24 '16 at 11:03

To have the answer in polar coordinates means to give it in the form of $re^{i\theta}$ (where $r$ is the radius and $\theta$ is the angle when drawing $re^{i\theta}$ on the complex plane).

In your case, you found (correctly) that $e^z = e^te^{i\varphi}$. Here $e^t$ plays the role of $r$ and $e^{i\varphi}$ plays the role of $e^{i\theta}$. So yes, the $e^{i\varphi}$ is part of the image of the map, and it tells you that your angle is $\varphi$.

Note that both the radius and the angle change as $t$ varies. More specifically, comparing $\varphi_0:=at_0+b$ and $\varphi_k:=a(t_0+k\frac{2\pi}{a})+b$, we see that these angles are equivalent since $$\varphi_k=at_0+b+k2\pi=\varphi_0+k2\pi$$ In other words, the image of the map keeps crossing each angle periodically. Combining this with a radius that increases as $t$ becomes big, this produces a spiral (which grows in a counterclockwise sense)


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