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In this question I learnt that the circumference of a unit circle in a complex plane is just the circumference of any normal circle $2\pi r$.

Now I would like to take into account the imaginary character of intervals on the imaginery axis. For this purpose I draw a circle in a complex plane, with the radius of one unit, and I get something that is no real circle because its radius is varying between 1 and i, depending on the direction of the radius. What would be the circumference of such a "circle"? I guess that it might be $\frac{\pi}{2} (1+i) $. Is my guess right?

Edit: The assumption of @Irregular User:

Similarly, if you measure the radius of your circle in the complex plane in the direction of y-axis positively increasing, then you see that your circle meets the y-axis at i. But the circle still has radius 1.

misunderstands my question, because I want to take into account the complex character of intervals on the imaginary axis. If I draw a circle on a complex plane I don't get a circle, because its horizontal radius is one, but its vertical radius is i.

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Moonraker
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    Radius is varying between Abs(1) = 1 and Abs(i) = 1 – Coolwater Jun 02 '16 at 20:46
  • It is a little unclear to me what you mean. The radius should not be "varying" since otherwise the resulting figure would not be a circle. If the radius is fixed at length $1$, it will remain the same as the angle changes. – M47145 Jun 02 '16 at 20:52

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If I'm interpreting your question correctly, you're saying that if you measure the radius of your circle in the complex plane in the direction of $x$-axis positively increasing, then you see that your circle meets the $x$-axis at $1$ and thus has radius $1$.

Similarly, if you measure the radius of your circle in the complex plane in the direction of $y$-axis positively increasing, then you see that your circle meets the $y$-axis at $i$. But the circle still has radius $1$.

Remember, when you look at the radius in this way, you must take the absolute value of what you've read off the axis. In a more familiar setting to you, when you look at a unit circle in the real plane, the circle also hits the $x$-axis and $y$-axis at $-1$ and $-1$. But this doesn't mean that the circle has a radius of $-1$. Again, we take the absolute value of $-1$, i.e. $|-1| = 1$.

Similarly, in our setting of the complex plane, $|1| = |-1| = |i| = |-i| = 1$.

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