Questions about the arg of complex numbers

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  • #1
Santiago24
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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
{\displaystyle \arg {z}=\{\operatorname {Arg} z+2\pi n:n\in \mathbb {Z} \}}
, i can't understand where come this and why there is some random integer, i understand
{\displaystyle \arg(z)=\phi \,}
but 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.
 

Answers and Replies

  • #2
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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
{\displaystyle \arg {z}=\{\operatorname {Arg} z+2\pi n:n\in \mathbb {Z} \}}
,
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##.
Santiago Perini said:
i can't understand where come this and why there is some random integer, i understand
{\displaystyle \arg(z)=\phi \,}
but 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.
Because ##8i## is an imaginary number that makes an angle of ##\frac \pi 2## with the real axis.
 
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  • #3
Santiago24
32
6
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.
Thanks for answer your comment cleared my doubts!
 
  • #4
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Here you find something about modulus and argument of complex numbers.

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)
 
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Likes Santiago24 and jedishrfu
  • #5
Santiago24
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Here you find something about modulus and argument of complex numbers.

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

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