- #1

Santiago24

- 32

- 6

z^3 = 8i

3 arg z = arg(z^3) = arg(8i ) = π/2 + 2kπ,

my question about this is where he find π/2.

If someone can answer and clear my doubts i will be very thankful.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Santiago24
- Start date

- #1

Santiago24

- 32

- 6

z^3 = 8i

3 arg z = arg(z^3) = arg(8i ) = π/2 + 2kπ,

my question about this is where he find π/2.

If someone can answer and clear my doubts i will be very thankful.

- #2

Mark44

Mentor

- 36,708

- 8,702

In the formula above ##\arg z## and ##\text{Arg} z## represent, respectively, the angle that z makes with the horizontal (i.e., real) axis, and the angle in the interval ##[0, 2\pi)##.Hi PF community, I'm reading about complex numbers and i have some questions about the argument of a complex number that i can't solve with Google or reading again the same page. Well, my first doubt is about,

In your example below, ##\text{Arg} z= \frac \pi 2## since that's the angle that the imaginary number ##i## makes with the real axis. Keep in mind that ##\frac \pi 2, \frac {5\pi} 2, \frac{9\pi} 2## and so on, are all possible angles for ##i##. Each of these is ##\frac \pi 2## plus some integer multiple of ##2\pi##. All of these angles are possible values for ##\arg z## when ##\text{Arg} z= \frac \pi 2##.

Because ##8i## is an imaginary number that makes an angle of ##\frac \pi 2## with the real axis.Santiago Perini said:i can't understand where come this and why there is some random integer, i understandbut the previous one is impossible to me. My other doubt is about an example that i saw.

z^3 = 8i

3 arg z = arg(z^3) = arg(8i ) = π/2 + 2kπ,

my question about this is where he find π/2.

If someone can answer and clear my doubts i will be very thankful.

- #3

Santiago24

- 32

- 6

Thanks for answer your comment cleared my doubts!In the formula above ##\arg z## and ##\text{Arg} z## represent, respectively, the angle that z makes with the horizontal (i.e., real) axis, and the angle in the interval ##[0, 2\pi)##.

In your example below, ##\text{Arg} z= \frac \pi 2## since that's the angle that the imaginary number ##i## makes with the real axis. Keep in mind that ##\frac \pi 2, \frac {5\pi} 2, \frac{9\pi} 2## and so on, are all possible angles for ##i##. Each of these is ##\frac \pi 2## plus some integer multiple of ##2\pi##. All of these angles are possible values for ##\arg z## when ##\text{Arg} z= \frac \pi 2##.

Because ##8i## is an imaginary number that makes an angle of ##\frac \pi 2## with the real axis.

- #4

BvU

Science Advisor

Homework Helper

- 15,272

- 4,252

The snag in your example is that ##z^3 = 8i## has several solutions, so just dividing the principal value, sometimes denoted by ##\operatorname {Arg}(8i)## by 3 is not sufficient and one has to look at the full set ##\operatorname {Arg}(8i)+ 2k\pi##. So in the range ##(-\pi,\pi]## one third of that gives us ##{1\over 6}\pi ##, ##{5\over 6}\pi ## and ##-{1\over 2}\pi##, namely with (##k = 0, 1, -1##, respectively)

- #5

Santiago24

- 32

- 6

Thanks for answer! and i'll check the link. :D

The snag in your example is that ##z^3 = 8i## has several solutions, so just dividing the principal value, sometimes denoted by ##\operatorname {Arg}(8i)## by 3 is not sufficient and one has to look at the full set ##\operatorname {Arg}(8i)+ 2k\pi##. So in the range ##(-\pi,\pi]## one third of that gives us ##{1\over 6}\pi ##, ##{5\over 6}\pi ## and ##-{1\over 2}\pi##, namely with (##k = 0, 1, -1##, respectively)

Share:

- Replies
- 5

- Views
- 108

- Replies
- 1

- Views
- 364

- Last Post

- Replies
- 8

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 487

- Replies
- 5

- Views
- 635

- Last Post

- Replies
- 2

- Views
- 125

- Last Post

- Replies
- 4

- Views
- 635

- Last Post

- Replies
- 9

- Views
- 976

- Last Post

- Replies
- 28

- Views
- 994

- Last Post

- Replies
- 29

- Views
- 721