Questions about the arg of complex numbers

[Santiago24]

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

, i can't understand where come this and why there is some random integer, i understand

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.

 

[Mark44]

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 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 $\text{Arg} z= \frac \pi 2$.

i can't understand where come this and why there is some random integer, i understand

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.

 

[Santiago24]

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!

 

[BvU]

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)

 

[Santiago24]

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