Find the modulus and the argument of ##\dfrac{2}{(4-2i)^2}##

[chwala]

Homework Statement

See attached- The question is set by me.

Relevant Equations

Complex Numbers

In my lines i have,

$(4-2i)^2 = (4-2i)(4-2i)$
$r^2 = 4^2 + (-2)^2 = 20$
$r \cos θ = 4$ and $r\sin θ = -2$
$\tan θ =-\dfrac{1}{2}$
$⇒θ = 5.82$ radians.

Therefore,

$|(4-2i)^2| = \sqrt{20} ×\sqrt{20} = 20$

Argument = $5.82 + 5.82 = 11.64$.

also $|2|$ = $2$ and argument = $0$.

Therefore,

$\left|\dfrac{2}{(4-2i)^2}\right| = \dfrac{2}{20}=\dfrac{1}{10}$

Argument of $\dfrac{2}{(4-2i)^2}=0 -11.64 = -11.64$ radians.

Insight welcome.

 

[MatinSAR]

First of all, I agree with your answer.

Another way is that you square the denominator. Then you divide the numerator by denominator. Now you have a complex number in form of $a+ib$ and you know how to find modulus and argument of it.
$$\dfrac {2}{(4-2i)^2}=\dfrac {2}{12-16i}=\dfrac {2}{12-16i} \dfrac {12+16i}{12+16i}=\dfrac {24+32i}{400}=0.06+0.08i$$
Edit: You don't need calculator until the last part for calculating the argument.
$$(12-16i)(12+16i)=12^2+16^2=16(3^2+4^2)=16(25)=400$$

So there is no need to use calculator.

 

[chwala]

First of all, I agree with your answer.

Another way is that you square the denominator. Then you divide the numerator by denominator. Now you have a complex number in form of $a+ib$ and you know how to find modulus and argument of it.
$$\dfrac {2}{(4-2i)^2}=\dfrac {2}{12-16i}=\dfrac {2}{12-16i} \dfrac {12+16i}{12+16i}=\dfrac {24+32i}{400}=0.06+0.08i$$
Edit: You don't need calculator until the last part for calculating the argument.
$$(12-16i)(12+16i)=12^2+16^2=16(3^2+4^2)=16(25)=400$$
View attachment 342157
So there is no need to use calculator.

Thanks mate...Math is diverse... refreshing is of essence at all times...complex numbers for today.

 

[PeroK]

I would have done it this way:
$$z = \frac 2{(4-2i)^2} = \frac{2(4+2i)^2}{20^2} = \frac{24 +32i}{400} = \frac{3+4i}{50} = \frac 1{10}(\frac 3 5 + i\frac 4 5)$$So that $|z| = \frac 1 {10}$ and $\arg(z) =\cos^{-1}(3/5) \approx 0.927$.

Argument of $\dfrac{2}{(4-2i)^2}=0 -11.64 = -11.64$ radians.

Insight welcome.

That's a fairly bizarre answer, as it's the principal argument minus $4\pi$.

 

[PeroK]

Note also that:
$$\bigg |\frac 2 {(a-ib)^2}\bigg | = \frac 2{|(a - ib)|^2} = \frac 2 {a^2 + b^2}$$And
$$\arg \bigg(\frac 2 {(a-ib)^2} \bigg) = \arg\big ((a+ib)^2\big ) = 2\arg(a+ib)$$So, in this case:
$$\arg(z) = 2\tan^{-1}(1/2) \approx 0.927$$