- #1
nickolas2730
- 28
- 0
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
limz→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,
limz→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!
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
limz→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,
limz→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!