Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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?

share|improve this answer
    
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.

share|improve this answer
    
Very nice your observation on point $a)$ –  user100106 Dec 31 '13 at 2:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.