# Complex Numbers and Circle

I am having a problem with the following question. They ask me for what values of a+bi is $e^{(a+ib)t}$ a circle. We have $t \in \mathbb{R}$

I think that since the modulus of $e^{(a+ib)t}$ is $e^{at}$ we must have a=0. And if a=0, we can then take any value of b (ie $b\in \mathbb{R}$-{0})

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If $t$ should vary between $0$ and $1$ you can't take any $b$. (You may just get a part of a circle.) – mrf Feb 18 '13 at 15:14
Sorry. I made a mistake while typing. t is in R. So $b \in \mathbb{R}$-{0} right? – user43418 Feb 18 '13 at 15:16
Is that correct? – user43418 Feb 18 '13 at 15:20

Your reasoning is totally correct. $e^{(a+bi)t}$ is a circle if and only if $|e^{(a+bi)t}|=e^{at}$ is constant as $t$ varies. Hence $a=0$ and $b\in \mathbb{R}$.
I have just one more question. How would you describe the path of $e^{(a+ib)t}$ if a>0? ie how does the point move as t increases – user43418 Feb 18 '13 at 15:14
You are right again. More precisely, outward spiral if $a>0$, inward spiral if $a<0$, both counterclockwise. – user54147 Feb 18 '13 at 15:25
Since $t$ is running through all real numbers, the value of $b$ makes no difference in this case as long as it is Real. – user54147 Feb 18 '13 at 15:34