complex z, lzl<1, map z to e^z, how many connected parts

[cathy_liping]

i know complex z is a unit circle, and when z is real axis, this transformation will map z to e^x, i think this is one connected part.

but what about if z is imaginary axis, what is the transformation through e^z

thank you in advance!

[owlpride]

One?

The exponential function is continuous, the open disk |z|<1 is connected, and the continuous image of a connected set is connected.

[Topoltergeist]

Here are my musings on the subject. I hope you find it useful.

The exponential map on complex numbers is similar to the Cartesian-polar change of coordinates in 2D: $$e^{x+iy} = e^x(\cos y + i \sin y).$$ Horizontal lines such as $$z(x) = x + bi$$ will be mapped to rays extending from the origin, with the complex component $$b$$ determining the angle of inclination. Vertical lines such as $$z(y) = a + i y$$ will be mapped to circles centered at the origin. The real component, $$a$$ will determine the radius of the circle. The imaginary axis gets mapped to the unit circle. (In fact, the exponential map is a covering map of the circle from a universal cover.)

[cathy_liping]

Here are my musings on the subject. I hope you find it useful.

The exponential map on complex numbers is similar to the Cartesian-polar change of coordinates in 2D: $$e^{x+iy} = e^x(\cos y + i \sin y).$$ Horizontal lines such as $$z(x) = x + bi$$ will be mapped to rays extending from the origin, with the complex component $$b$$ determining the angle of inclination. Vertical lines such as $$z(y) = a + i y$$ will be mapped to circles centered at the origin. The real component, $$a$$ will determine the radius of the circle. The imaginary axis gets mapped to the unit circle. (In fact, the exponential map is a covering map of the circle from a universal cover.)

thank you for your answer first! i think the real axis and imaginary axis are two connected components. and what about other parts except the real axis and imaginary axis?