Find the modulus of ##\dfrac{(2-3i)(3+4i)}{(6+4i)(16-8i)}##

[chwala]

Homework Statement

Find the modulus of $\dfrac{(2-3i)(3+4i)}{(6+4i)(16-8i)}$

Relevant Equations

Complex Numbers

In my working i have,

$\dfrac{\sqrt 13\sqrt 25}{\sqrt 52\sqrt 320}=\dfrac{\sqrt 13⋅5}{\sqrt 52⋅8\sqrt 5}=\dfrac{\sqrt 1⋅5⋅\sqrt 5}{\sqrt 4⋅\sqrt 5⋅8\sqrt 5}=\dfrac{1⋅\sqrt 5}{2⋅8}=\dfrac{\sqrt 5}{16}$

The textbook says otherwise... textbook answer $\dfrac{5}{34}$.

 

[martinbn]

Their answer is wrong, but that doesn't mean yours is ok. What happened to $\sqrt{13}$?

 

[Gavran]

$\dfrac{\sqrt 13\sqrt 25}{\sqrt 52\sqrt 320}=\dfrac{\sqrt 13⋅5}{\sqrt 52⋅8\sqrt 5}=\dfrac{\sqrt 1⋅5⋅\sqrt 5}{\sqrt 4⋅\sqrt 5⋅8\sqrt 5}=\dfrac{1⋅\sqrt 5}{2⋅8}=\dfrac{\sqrt 5}{16}$

Okay.

 

[chwala]

Their answer is wrong, but that doesn't mean yours is ok. What happened to $\sqrt{13}$?

@martinbn ... just check my working again. Cheers man.

 

[Mark44]

What happened to $\sqrt{13}$?

It cancelled with the factor in the denominator since 52 = (4)(13).
$\frac {\sqrt{13}}{\sqrt{52}} = \frac 1 2$

 

[Mark44]

The textbook says otherwise

Is the problem you posted exactly the same as the one in the textbook?

 

[chwala]

Is the problem you posted exactly the same as the one in the textbook?

Yes @Mark44

 

[Mark44]

Find the modulus of $\dfrac{(2-3i)(3+4i)}{(6+4i)(16-8i)}$

Since you're often interested in different approaches, here's another way to go:
Multiply the pairs of complex numbers in the numerator and the denominator to get $\frac{18 - i}{128 + 16i} = \frac{18 - i}{16(8 + i)}$.
Now take the modulus (or magnitude) of the numerator and denominator to get $\frac{\sqrt 5}{16}$, the same as you got using your method.
In terms of ease of calculation, I like your way better, but doing things a different way and getting the same result is more evidence that the textbook answer is incorrect.

 

[SammyS]

@chwala
Quoting Mark:

Since you're often interested in different approaches, here's another way to go:

Split the given expression into the product of two complex fractions.

Noting that $\displaystyle \left | 6+4i \right | =2 \left | 3+2i \right |$ and that $\displaystyle \left | 3+2i \right | =\left | 2-3i \right |$ ,

you have $\displaystyle \left | \dfrac {2-3i}{6+4i} \right | = \dfrac 1 2$

The other fraction:
Factor out 8 from the denominator: $\displaystyle 16-8i =8 \left ( 2-i \right )$

Familiarity with the $3,\,4,\,5\,$ right triangle gives $\displaystyle \left | 3+4i \right | =5$.

Finally, calculation gives: $\displaystyle \left | 2-i \right | = \sqrt {5\,}$ .

Put this all together to get:

$\displaystyle \quad \quad \left | \dfrac {(2-3i)(3+4i)}{(6+4i)(16-8i)} \right | = \dfrac 1 2 \cdot \dfrac { 5}{ 8\,\sqrt{5\,}} = \dfrac { \sqrt{5\,}}{ 16}$