What does the graph of $5e^{it}$ look like on the complex plane?

Question

I know that $5e^{it} = 5(\cos(t) + i\sin(t))$, but that doesn't really help me. What other information can I use to visualize this graph besides plotting many points and seeing what type of graph it gives me?

Answer 1

Perhaps you really mean the image of the function $f:\mathbb R\to \mathbb C$ with $f(t)=5e^{it}$? It lies in $\mathbb C$, and is a circle of radius $5$ centered at $0$.

That's distinct from the graph of the function, which lies in $\mathbb R\times \mathbb C$.

Addendum 1: The graph is actually easy to visualize as well. As $t$ increases through $\mathbb R$, the image point $f(t) =5e^{it}$ moves around the image circle counterclockwise. So the graph is a spiral. If you think of the range $\mathbb C$ as $\mathbb R\times \mathbb R$, identifying $x+iy \leftrightarrow (x,y)$, and thinking of the graph as lying in $\mathbb R\times (\mathbb R\times \mathbb R)\simeq\mathbb R\times \mathbb R\times \mathbb R$, the points on the graph are $(t,5\cos t, 5\sin t)$; they form a counterclockwise spiral of radius $5$ around the $x$-axis. The image of the function is the projection of the graph onto the $yz$-plane, a circle.

Addendum 2: Here's a plot of the graph:

Answer 2

It is a circle around the origin having radius 5 if you draw it as 2D points $(x(t), y(t)) = (5 \cos t, 5 \sin t)$.

Check it here.