### Author Topic: About the definition of Argument (in book)  (Read 3134 times)

#### Ende Jin

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##### About the definition of Argument (in book)
« on: September 08, 2018, 03:39:53 PM »
I found that the definition of "arg" and "Arg" in the book is different from that introduced in the lecture (exactly opposite) (on page 7).
I remember in the lecture, the "arg" is the one always lies in $(-\pi, \pi]$
Which one should I use?

#### Victor Ivrii

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##### Re: About the definition of Argument (in book)
« Reply #1 on: September 08, 2018, 04:58:09 PM »
Quote
Which one should I use?
This is a good and tricky question because the answer is nuanced:
Solving problems, use definition as in the Textbook, unless the problem under consideration requires modification: for example, if we are restricted to the right half-plane  $\{z\colon \Re z >0\}$ then it is reasonable to consider $\arg z\in (-\pi/2,\pi/2)$, but if we are restricted to the upper half-plane  $\{z\colon \Im z >0\}$ then it is reasonable to consider $\arg z\in (0,\pi)$ and so on.

#### Ende Jin

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##### Re: About the definition of Argument (in book)
« Reply #2 on: September 09, 2018, 12:48:34 PM »
I am still confused. Let me rephrase the question again.
In the textbook, the definition of "arg" and "Arg" are:
$arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = cos\theta + isin\theta$
which means $arg(z) \in \mathbb{R}$
while
$Arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = cos\theta + isin\theta \land \theta \in [-\pi, \pi)$
which means $Arg(z) \in [-\pi, \pi)$

While in the lecture, as you have introduced, it is the opposite and the range changes to $(-\pi, \pi]$ instead of $[-\pi, \pi)$ (unless I remember incorrectly):
Arg is defined to be
$Arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = (cos\theta + isin\theta)$
which means $arg(z) \in \mathbb{R}$
while arg is
$arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = cos\theta + isin\theta \land \theta \in (-\pi, \pi]$

I am confused because if I am using the definition by the book,
when $z \in \{z : Re (z) > 0\}$
then $arg(z) \in (-\frac{\pi}{2} + 2\pi n,\frac{\pi}{2} + 2\pi n), n \in \mathbb{Z}$

#### Victor Ivrii

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##### Re: About the definition of Argument (in book)
« Reply #3 on: September 09, 2018, 04:40:38 PM »

BTW, you need to write \sin t and \cos t and so on to have them displayed properly (upright and with a space after): $\sin t$, $\cos t$ and so on
« Last Edit: September 12, 2018, 04:38:06 PM by Victor Ivrii »

#### Ende Jin

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##### Re: About the definition of Argument (in book)
« Reply #4 on: September 10, 2018, 10:03:33 AM »
Thus in a test/quiz/exam, I should follow the convention of the textbook, right?

#### Victor Ivrii

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##### Re: About the definition of Argument (in book)
« Reply #5 on: September 10, 2018, 01:34:22 PM »
Thus in a test/quiz/exam, I should follow the convention of the textbook, right?
Indeed

#### oighea

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##### Re: About the definition of Argument (in book)
« Reply #6 on: September 12, 2018, 04:24:46 PM »
The $\arg$ of a complex number $z$ is an angle $\theta$. All angles $\theta$ have an infinite number of "equivalent" angles, namely $\theta =2k\pi$ for any integer $k$.

Equivalent angles can be characterized by that they exactly overlap when graphed on a graph paper, relative to the $0^\circ$ mark (usually the positive $x$-axis). Or more mathematically, they have the same sine and cosine. It also makes sine and cosine a non-reversible function, as given a sine or cosine, there are an infinite number of angles that satisfy this property.

$\Arg$, on the other hand, reduces the range of the possible angles such that it always lie between $0$ (inclusive) to $2\pi$ (exclusive). That is because one revolution is $2\pi$, or $360$ degrees. That is called the principal argument of a complex number.

We will later discover that complex logarithm also have a similar phenomenon.
« Last Edit: September 12, 2018, 04:37:32 PM by Victor Ivrii »