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I have to calculate the absolute value of $\lvert{i(2+3i)(5-2i)\over(2-3i)^3}\rvert$ solely using properties of modulus, not actually calculating the answer. I know that I could take the absolute value of the numerator and denominator and then take the absolute value of each factor, giving me $\lvert i \rvert \lvert2+3i\rvert\lvert5-2i\rvert\over \lvert(2-3i)^3\rvert$. I'm not sure what to do after this, without calculating it. Using $\lvert z \rvert=\sqrt{x^2 + y^2}$ I can plug in each fator into the equation and simplify but that would be calculating it, and I'm not sure if I'm suppose to go that far. Help please.

cele
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  • Every complex number can be written in the form $a+bi$, simplify the number into that form and apply the modulus? What do you mean by solely using properties of the modulus? – JessicaK Oct 04 '14 at 23:23
  • Properties like $\vert z \rvert$ above and $\lvert {z_1\over z_2}\evert$ – cele Oct 04 '14 at 23:29

3 Answers3

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$\frac{|i(2i+3)(5-2i|}{|(2-3i)^3|}=\frac{|i||2i+3||5-2i|}{|2-3i|^3}=\frac{\sqrt{0^2+1^2}\sqrt{2^2+3^2}\sqrt{5^2+(-2)^2}}{(\sqrt{2^2+(-3)^2})^3}=\frac{1\cdot\sqrt{13}\sqrt{29}}{(\sqrt{13})^3}=\frac{\sqrt{29}}{13}$ Notice that we have taken the positive values out of the square roots as the modulus of a complex number is a non-negative real number.

Arian
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  • Thanks, that's what I got too but I was told to not calculate so I wasn't sure how far to go with it – cele Oct 04 '14 at 23:36
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It seems to me that at some point one is going to want to use

$\vert a + bi \vert^2 = a^2 + b^2; \tag{1}$

that is, if an answer expressed as a single, non-negative real number is the ultimate goal. What we can do, however, is to carry things as far as we possible without employing (1), thus minimizing (hopefully) the amount of arithmetic to be done; we do this by working with more abstract properties of the modulus, such as

$\vert z_1 z_2 \vert = \vert z_1 \vert \vert z_2 \vert \tag{2}$

and

$\vert z \vert = \vert \bar z \vert\, \tag{3}$

etc. Bearing this intention in mind, we may proceed. Our OP cele has already taken a solid first step with

$\vert \dfrac{(i)(2 + 3i)(5 -2i)}{(2- 3i)^3} \vert = \dfrac{\vert i \vert \vert 2 + 3i \vert \vert 5 - 2i \vert}{\vert (2 - 3i)^3 \vert}; \tag{4}$

proceeding from ($4$), we note that several simplifications can be made before we actually invoke ($1$). For example, $\vert i \vert = 1$; this of course follows almost trivially from ($1$), but we also note we might observe that $\vert 1 \vert = 1$ by virtue of ($2$): $\vert 1 \vert = \vert 1^2 \vert = \vert 1 \vert^2$, whence $\vert 1 \vert \ne 0$ implies $\vert 1 \vert = 1$; from this logic we have $\vert -1 \vert = 1$ as well, since $1 = \vert 1 \vert = \vert (-1)^2 \vert = \vert -1 \vert^2$, since we must have $\vert -1 \vert > 0$, $\vert -1 \vert = 1$; then

$\vert i \vert^2 = \vert i^2 \vert = \vert -1 \vert = 1, \tag{5}$

so

$\vert i \vert = 1; \tag{6}$

$\vert -i \vert = 1$ follows similarly. We thus deduce, without invoking ($1$), that the factor $\vert i \vert$ in the numerator of the right-hand side of ($4$) is of no consequence; we may ignore it. Turning next to the factors $\vert 2 + 3i \vert$ and $\vert (2 - 3i)^2 \vert$, we have, again from ($2$), that

$\vert (2 - 3i)^3 \vert = \vert 2 - 3i \vert^3 \tag{7}$

and from ($3$)

$\vert 2 + 3i \vert = \vert 2 - 3i \vert; \tag{8}$

(6), (7) and (8) lead to the conclusion that

$\dfrac{\vert i \vert \vert 2 + 3i \vert \vert 5 - 2i \vert}{\vert (2 - 3i)^3 \vert} = \dfrac{\vert 5 - 2i \vert}{\vert 2 - 3i \vert^2}. \tag{9}$

I can't see how to take this any further without ($1$); using it yields

$\vert \dfrac{(i)(2 + 3i)(5 -2i)}{(2- 3i)^3} \vert = \dfrac{\vert i \vert \vert 2 + 3i \vert \vert 5 - 2i \vert}{\vert (2 - 3i)^3 \vert} = \dfrac{\sqrt{29}}{13}, \tag{10}$

in agreeement with Adrian's answer.

I think one point of this exercise is to illustrate the differences encountered when considering $\vert z \vert$ as a homomorphism from the group of non-zero complex numbers $\Bbb C^\ast$ to the positive reals $\Bbb R_+^\ast$ versus a norm on the vector space $\Bbb C$. In the former case start with "axioms" like (2) and (3) and then deduce algebraic facts about $\vert z \vert$; in the latter we start with (1) and prove things like (2)-(3). Ultimately, of course, the two views are equivalent, but they often appear different in practice.

Finally, it would be interesting to know if a homomorphism $\phi: \Bbb C^\ast \to \Bbb R^\ast$ ($2$) which is conjugate-invariant, that is $\phi(z) = \phi(\bar z)$ ($3$), (I think "$\phi$ factors through the conjugation involution" is the way the group theorists like to say it?) shares many properties in common with $\vert \cdot \vert$ as defined by (1). Do we need more axioms than (2) and (3) to attain the equivalence? I'd be glad to hear from anyone who knows.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Robert Lewis
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  • I'm pretty sure that the only continuous homomorphisms $\phi:\mathbb{C}^{\times} \rightarrow \mathbb{R}^{\times}$ are of the form $\phi(z) = |z|^s$ for some real number $s$. If you drop the continuity requirement there are most likely pathological examples. – Alex Zorn Oct 05 '14 at 04:53
  • @AlexZorn: I'm pretty sure that's right; I worked out a bunch of a proof; if I get it all, I'll probably post it as as self-answered question. Thanks for the input! – Robert Lewis Oct 05 '14 at 04:58
  • @AlexZorn: By the way, are all continuous homomorphisms $\Bbb R^\ast \to \Bbb R^\ast$ of the form $x^s$ for fixed real $s \ne 0$? – Robert Lewis Oct 05 '14 at 05:00
  • If you mean $\mathbb{R}{> 0} \rightarrow \mathbb{R}{> 0}$, then yes. For $\mathbb{R}^{\times} \rightarrow \mathbb{R}^{\times}$ you have $x \mapsto |x|^{s}$, but you also have $sgn(x)$, which sends any positive real number to $1$ and any negative real number to $-1$. And then you also get $x \mapsto sgn(x)|x|^{s}$ for any real number $s$. – Alex Zorn Oct 06 '14 at 21:19
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This question is ancient, so I'm not sure why I'm bothering, but I'm ~100% sure that a typo in the original question is why this seems complicated.

The $5+2i$ should be a $5+12i$ :

$$\left|i(2+3i)(5+12i) \over (2 - 3i)^3\right|$$

So the simplification is just $$\begin{align} \left|i(2+3i)(5+12i) \over (2 - 3i)^3\right| &= {|i||2+3i||5+12i| \over |2-3i||2-3i|^2}\cr &= |i| {|2 + 3i| \over |2 - 3i|}{|5+12i| \over |2-3i|^2}\cr &= (1)(1)(1)\cr &= 1 \end{align} $$

user295691
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