Find and sketch the image of the straight line $z = (1+ia)t+ib$ under the map $w=e^{z}$

Question

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!

Answer 1

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.

Answer 2

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)