Question

heres the question it deals with complex number yet I dont know exactly whats its asking.

Let H(\omega) be a complex-valued function of the real variable \omega. Find |H(\omega)| and argH(\omega) equations below

Answer 1

\[H(\omega) = \frac{ 1 }{ (1+i\omega)^{10}}\]

Answer 2

i think you need to convert it to polar form
r(cosx +isinx)

where r = modulus and x is the argument

Answer 3

ok so \[\left| H(\omega) \right|\] is r?

Answer 4

yes, but i'm not sure if you have to work it out

Answer 5

i'm currently working out the argument, and i think i've got it too

Answer 6

ok

Answer 7

there's several rules for arguments

\[\large \arg(z^n)=n~\arg(z)\]
\[\large \arg(\frac{ z_1 }{ z_2 })=\arg(z_1)-\arg(z_2)\]
\[\large \arg(x+yi)=\tan^{-1}(\frac{ y }{ x })\]

Answer 8

if i'm not mistaken that would mean we have arg(1)-10 arg(1+i\(\omega\))

Answer 9

which would be 0-tan\(^{-1}(\omega)\) or simply -tan\(^{-1}(\omega)\)

Answer 10

ok Im trying to follow.

Answer 11

actually i forgot the 10 in the last one, so \(-10tan^{-1}(\omega)\)

Answer 12

Let \(z=1+i\omega\), then
\[\begin{cases}
r\cos\theta=1\\
r\sin\theta=\omega
\end{cases}~~\implies~~r^2\cos^2\theta+r^2\sin^2\theta=1+\omega^2=r^2\]
so \(r=\sqrt{1+\omega^2}\).
\[\theta=\tan^{-1}\frac{y}{x}=\tan^{-1}\omega\]
So you have
\[\begin{align*}H(\omega)&=\frac{1}{(1+i\omega)^{10}}\\\\
&=\frac{1}{\left(\sqrt{1+\omega^2}\exp\left(i\tan^{-1}\omega\right)\right)^{10}}\\\\
&=\frac{1}{\left(1+\omega^2\right)^5\exp\left(10i\tan^{-1}\omega\right)}&\text{ by DeMoivre's theorem}
\end{align*}\]
Try taking the modulus now.

Answer 13

thanks guys!