Questions about the arg of complex numbers

In summary, the conversation discusses complex numbers and the argument of a complex number. The formula for the argument is explained, and an example is given where the argument is equal to π/2. It is noted that there can be multiple solutions for a given complex number and the full set must be considered. The conversation concludes with a thank you and a reference to further reading on the topic.
  • #1
Santiago24
32
6
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.
 
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  • #2
Santiago Perini said:
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
Mark44 said:
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
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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  • #5
BvU said:
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
 

Related to Questions about the arg of complex numbers

1. What is the definition of the argument of a complex number?

The argument of a complex number is the angle formed between the positive real axis and the vector representing the complex number in the complex plane. It is typically measured in radians or degrees.

2. How is the argument of a complex number calculated?

The argument of a complex number can be calculated using the inverse tangent function (arctan) of the imaginary part divided by the real part. It is also possible to calculate the argument using the complex conjugate of the complex number.

3. What is the range of possible values for the argument of a complex number?

The range of possible values for the argument of a complex number is between -π (inclusive) and π (exclusive). This range represents a full rotation around the complex plane.

4. How does the argument of a complex number relate to its magnitude?

The argument of a complex number and its magnitude (or absolute value) are two different properties. The argument represents the direction of the complex number in the complex plane, while the magnitude represents its distance from the origin.

5. Can the argument of a complex number be negative?

Yes, the argument of a complex number can be negative. A negative argument represents a direction in the opposite direction of the positive real axis, or a rotation in the clockwise direction in the complex plane.

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