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I am working on basic problems about complex numbers and functions in order to learn complex analysis from Lang's textbook. I am trying to solve the following exercise:

Let $f:\mathbb C \to \mathbb C$, $f(z)=e^z$ (if $z=a+bi$, then $e^z=e^ae^{ib}$)

a) Find the image under $f$ of the set $S=\{z \in \mathbb C : 0\leq Im(z)<2\pi\}$.

b) Show that the image of the line $\{t+it : t \in \mathbb R\}$ is a spiral.

For a), I am not so sure how to describe the image, I know that if $z=a+bi$, then $b \in [0,2\pi)$. By definition, $e^{ib}=\cos(b)+i\sin(b)$, so the image of the set $S$ is $T=\{w \in \mathbb C : w=f(z)=e^a(\cos(b)+i\sin(b)), \space b \in [0,2\pi)\}$

Is this a correct way of describing the image of the given set?

For part b), I don't know how to show this, maybe I need to find a possible parameterization $\phi(t)$ of the set $\{(u(t),v(t),t) \in \mathbb R^3\}$ where if $w=f(z(t))$, then $w=u(t)+iv(t)$.

If $z$ is in the line, then $z=t+it$, and $f(z)=e^t(\cos(t)+i\sin(t))=e^t\cos(t)+ie^t\sin(t)$,

so the parametrization would be $\phi(t)=(e^t\cos(t),e^t\sin(t),t)$

Is this the parameterization of a spiral?

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up vote 2 down vote accepted

a) So far perfect. Observe that your formula already describes all points in the complex plane (use polar coordinates), except $0$.

b) Yes, the curve $t\mapsto e^{(1+i)t}$ is called an exponential or logarithmic spiral. In polar coordinates, the point at moment $t$ has length $e^t$ (distance from origin) and angle $t$. So, as angle increases, the length increases exponentially.

Where does the exponential map map the horizontal and vertical lines, by the way?

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I hope it's not considered bad etiquette if I too share my post, as I almost finished it when you posted yours. To 1) the image does not contain $0$. – benh Dec 31 '13 at 2:36
No, of course not. And, "ah, yes"... : ) – Berci Dec 31 '13 at 2:37
I was going to say the same thing about a). Both posts were of great help, thanks guys! – user100106 Dec 31 '13 at 2:50
Btw, the function maps a vertical line $a+it$ to a circle of radius $e^a$ and a horizontal line $t+ib$ to two exponentials (for $b$ fixed) of the form $\cos(b)e^t$ and $\sin(b)e^t$ – user100106 Dec 31 '13 at 2:53

1.) Your characterization is correct, but there is an easier description of $T$. Have a closer look at the values of $f(z) = e^a(\cos(b)+i\sin(b))$: If you fix $a$, and vary $b \in [0,2 \pi)$ then the image of $(cos(b)+i\sin(b))$ is the unit circle. If you fix $b$, you can scale the number by a positive real number $e^a$, as $e^a$ is onto $\Bbb R^+$. So you can interpret $T$ as the union of all circles in the complex plane with center $0$ and arbitrary positive radius. What's that?

2) Yes, that's a spiral, but why do you need a $3$-dimensional real vector to describe it? If you want to describe it with a real vector, two variables are sufficient. Normally you describe spirals in polar coordinates $(r(t),\theta(t))$ dependent of a parameter $t$ where $\theta(t)$ is interpreted as the angle and $r(t)$ as the radius. So in your case $r(t) = e^t$ and $\theta(t) = t$, a logarithmic spiral.

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Very nice your observation on point $a)$ – user100106 Dec 31 '13 at 2:55

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