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.
Answer
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!!!
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