Limits of z Approaching i Along Unit Circle |z|=1

[nickolas2730]

1. Consider the limit limz→i(|z| + iArg(iz))


2. (i) What value does the limit approach as z approaches i along the
unit circle |z| = 1 in the first quadrant?
(ii) What value does the limit approach as z approaches i along the
unit circle |z| = 1 in the second quadrant?



3. solutions are as follow:
(i) Note that iz = −y + ix. When z is in the first quadrant, iz is in the second quadrant. Thus, we have
lim_z→i(|z| + iArg(iz)) = 1 + i∏
when z approaches i along the unit circle |z| = 1 in the first quadrant.
(ii)Similarly, when z approaches i along the unit circle |z| = 1 in the
second quadrant,
lim_z→i(|z| + iArg(iz)) = 1 − i∏

My question is (i) why is it in second quadrant when z is in first quadrant?
is it because the real part is now -y and imaginary part is x which -y is the x-axis and x is the y-axis?
(ii)Why Arg(iz)=∏, when it is in second quadrant?

Thanks!

 

[HallsofIvy]

1. Consider the limit limz→i(|z| + iArg(iz))


2. (i) What value does the limit approach as z approaches i along the
unit circle |z| = 1 in the first quadrant?
(ii) What value does the limit approach as z approaches i along the
unit circle |z| = 1 in the second quadrant?

3. solutions are as follow:
(i) Note that iz = −y + ix. When z is in the first quadrant, iz is in the second quadrant. Thus, we have
lim_z→i(|z| + iArg(iz)) = 1 + i∏
when z approaches i along the unit circle |z| = 1 in the first quadrant.
(ii)Similarly, when z approaches i along the unit circle |z| = 1 in the
second quadrant,
lim_z→i(|z| + iArg(iz)) = 1 − i∏

My question is (i) why is it in second quadrant when z is in first quadrant?
is it because the real part is now -y and imaginary part is x which -y is the x-axis and x is the y-axis?
(ii)Why Arg(iz)=∏, when it is in second quadrant?

Any z on the unit circle can be represented as $z= e^{i\theta}$ so $iz= e^{i\pi/2}e^{i\theta}= e^{i(\theta+ \pi/2}$. Multiplying by i "rotates" the number 90 degrees. When z is in the first quadrant, iz is in the second quadrant. When z is in the second quadrant iz is in the third quadrant.

Thanks!