Finding the modulus and argument of ##\dfrac{a}{(b±ci)^n}##

[chwala]

Homework Statement

Finding the modulus and argument of complex numbers of the form $\dfrac{a}{(b±ci)^n}$ where $a,b,c,n ∈ \mathbb{Z}$

Relevant Equations

complex numbers

Doing a refreshing on this, let me come up with my own example,

Finding the modulus and argument of $\dfrac{4}{(1-i)^3}$,

Considering the denominator,
Let $z=1-i, |z|=\sqrt 2→ z^3=\sqrt 2⋅ \sqrt 2⋅\sqrt 2 = 2\sqrt 2$, noting that $arg z_1z_2=arg z_1 + arg z_2$, then
argument $z= -\dfrac{π}{4}-\dfrac{π}{4}-\dfrac{π}{4}=-\dfrac{3π}{4}$.

Considering the numerator,
modulus=4, argument = 0.


Now having both numerator and denominator in mind, and noting that modulus,

$\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}$

$\left|\dfrac{4}{2\sqrt2}\right|=\dfrac{|4|}{|2\sqrt2|}=\dfrac{2\sqrt2}{2}=\sqrt 2$
argument $Z= \left[0-\dfrac{3π}{4}\right]=\dfrac{3π}{4}$.

The steps seem clear and quite easy, question is does this apply to all forms i.e $\dfrac{a}{(b±ci)^n}$? or are there cases where this fails.

 

[Mark44]

Doing a refreshing on this, let me come up with an example, Finding the modulus and argument of $\dfrac{4}{(1-i)^2}$,
Considering the denominator,
i let $z=1-i, |z|=\sqrt 2→ z^3=\sqrt 2⋅ \sqrt 2⋅\sqrt 2 = 2\sqrt 2$, noting that $arg z_1z_2=arg z_1 + arg z_2$

Why do you have $z^3$?
Isn't $\left|\frac 4 {(1 - i)^2}\right| = \frac 4 2 = 2$?
To find arg of your example, you could multiply out the $(1 - i)^2$ factor to get -2i, so then you have $\frac 4 {-2i}$ (assuming my mental arithmetic is good), then rationalize that to get a complex number in the form of m + ni. You can find the angle this makes in a straightforward manner.

 

[chwala]

Why do you have $z^3$?
Isn't $\left|\frac 4 {(1 - i)^2}\right| = \frac 4 2 = 2$?
To find arg of your example, you could multiply out the $(1 - i)^2$ factor to get -2i, so then you have $\frac 4 {-2i}$ (assuming my mental arithmetic is good), then rationalize that to get a complex number in the form of m + ni. You can find the angle this makes in a straightforward manner.

amended ...typo.

If you get my thinking, I am looking at, say, $\dfrac{100}{(13-15i)^{49}}$ large numbers, you cannot multiply the complex number by way of expanding. The approach I used may be faster, I stand to be corrected.

 

[Mark44]

From your revised post #1:

Considering the denominator, Let $z=1-i, |z|=\sqrt 2→ z^3=\sqrt 2⋅ \sqrt 2⋅\sqrt 2 = 2\sqrt 2$, noting that $arg z_1z_2=arg z_1 + arg z_2$, then
argument $z= -\dfrac{π}{4}-\dfrac{π}{4}-\dfrac{π}{4}=-\dfrac{3π}{4}$.

You'd have to convince me that this is a valid step; i.e., that, say $arg(\frac a {b + ci}) = arg(b + ci)$.

For example, is $arg(\frac 1 {1 + i}) = arg(1 + i)$?

 

[chwala]

From your revised post #1:
You'd have to convince me that this is a valid step; i.e., that, say $arg(\frac a {b + ci}) = arg(b + ci)$.

For example, is $arg(\frac 1 {1 + i}) = arg(1 + i)$?

this should be, $arg(\frac 1 {1 + i}) = arg1 - arg(1 + i)$.

Using,

$arg \left(\dfrac{z_1}{z_2}\right)=arg z_1 -arg z_2$.

 

[DaveE]

polar, dude... polar
Granted, I'm more engineer than mathematician. I'll always take the easy path.

 

[WWGD]

And $arg(z1+z2)\neq argz1+argz2.$. You need to consider the " overflow" issues.

 

[PeroK]

And $arg(z1+z2)\neq argz1+argz2.$. You need to consider the " overflow" issues.

Take $z1 =1$ and $z2 = i$.

Although, I assume you're going to say that's an "overflow" issue.

 

[Mark44]

And $arg(z1+z2)\neq argz1+argz2.$. You need to consider the " overflow" issues.

@chwala is not asserting this -- only that $arg(z_1z_2) = arg(z_1) + arg(z_2)$ or that $arg\left(\frac{z_1}{z_2}\right) = arg(z_1) - arg(z_2)$

 

[Ibix]

I agree with @DaveE. Use $z=|z|e^{i\mathrm{arg}(z)}$. Job done two or three lines later.

 

[chwala]

And $arg(z1+z2)\neq argz1+argz2.$. You need to consider the " overflow" issues.

Where is this line coming from? i do not think i stated this anywhere.

 

[chwala]

I agree with @DaveE. Use $z=|z|e^{i\mathrm{arg}(z)}$. Job done two or three lines later.

Interesting approach,

$1-i=\sqrt 2 e^{\dfrac{-πi}{4}}$

$(1-i)^3=2\sqrt 2 e^{\dfrac{-3πi}{4}}$

$\dfrac{4}{(1-i)^3}=\dfrac{4}{2\sqrt 2e^\dfrac{-3πi}{4} }={\sqrt 2e^\dfrac{3πi}{4} }$

 

[WWGD]

@chwala is not asserting this -- only that $arg(z_1z_2) = arg(z_1) + arg(z_2)$ or that $arg\left(\frac{z_1}{z_2}\right) = arg(z_1) - arg(z_2)$

Ok, my bad. But the issue with the overflow remains: $arg(z1z2)=arg(z1)+arg(z2)$ only if the sum is less than $2 \pi$.

 

[Gavran]

It would be good to answer the original question.
$$ z=\frac{a}{(b\pm ci)^n}=|z|e^{i\arg(z)} $$.
What are $|z|$ and $\arg(z)$ in the general case?

 

[chwala]

It would be good to answer the original question.
$$ z=\frac{a}{(b\pm ci)^n}=|z|e^{i\arg(z)} $$.
What are $|z|$ and $\arg(z)$ in the general case?

@Gavran i hope i am getting you right, are you asking a question? i thought i had already addressed the modulus and argument in posts $1$ and $12$.

 

[Mark44]

i thought i had already addressed the modulus and argument in posts 1 and 12.

No, not for the general expression. You worked on the example you gave in post #1 : $\dfrac 4 {(1 - i)^3}$.

 

[chwala]

No, not for the general expression. You worked on the example you gave in post #1 : $\dfrac 4 {(1 - i)^3}$.

aaaaaah okay,

in polar form, $b+ci=re^{iθ}$

$(b+ci)^n = r^n e^{inθ}$

...

$z=\dfrac{ae^{-inθ}}{r^n}=\dfrac{ae^{-in\tan^{-1}\left(\frac{c}{b}\right)}}{(b^2+c^2)^\frac{n}{2}}$

Modulus: $\dfrac{a}{(b^2+c^2)^\frac{n}{2}}$

Argument: ${-n\tan^{-1}\left(\dfrac{c}{b}\right)}$

 

[DaveE]

Good. I think that's a good place to stop. Everyone knows what you mean by $−n ~ tan^{−1}(c/b).$

But, a rather pedantic and nitpicking extension: Your solution could result in $\theta=-\frac{17\pi}{4}$ in some case. Most would rather think of this as $\theta=-\frac{\pi}{4}$. Also, the argument for $b, c >0$ isn't the same as for $b, c <0$, but $tan^{-1}(c/b)$ is the same.

So there is an issue with the principle branch of $tan^{−1}$, $arg(z)$, $Arg(z)$, etc.
https://math.libretexts.org/Bookshelves/Analysis/Complex_Variables_with_Applications_(Orloff)/01:_Complex_Algebra_and_the_Complex_Plane/1.09:_The_function_arg(z)#:~:text=The branch −π<arg(,branch of arg(z).