MHB Newbie here with question on Complex components

[THAGONZ]

Hello all,

I am new here and very new to trig. I can't seem to figure out how this equation works out. I was wondering if someone could help me out with the long equation to get this answer. I know this is probably very simple to most of you but I am really stuck here...

Thanks in advance!(cos-isin)*(cos-isin)

(0.00179-i0.00358)*(0.29-i0.957) = -0.003-i0.003

 

[ineedhelpnow]

hi! (Wave) welcome to mhb. like your username.

was that all you were given in the question?

 

[THAGONZ]

hi! (Wave) welcome to mhb. like your username.

was that all you were given in the question?

Hi, Thank you for the reply..

It is actually a much longer equation...

$0.004[cos(-63.4^{\circ})+isin(-63.4^{\circ})]$=0.00179-i0.00358

then using Discrete Fourier Transform you get
$e-i2pi(29.84*10^6)*(1/5.12*10^6)*-7$= 0.29 – i0.957

so now I am trying to figure out how $(0.00179-i0.00358)*(0.29-i0.957)$ = -0.003-i0.003

 

[ineedhelpnow]

but a dollar sign at the beginning of your equations and at the end. I am having some trouble reading it.

 

[Opalg]

so now I am trying to figure out how $(0.00179-i0.00358)*(0.29-i0.957)$ = -0.003-i0.003

Multiply out the brackets in the usual way (remembering that $i^2=-1$) and then collect the real and imaginary terms together. That gives $$(0.00179-i0.00358)*(0.29-i0.957) = 0.00179*0.29 + 0.00358*0.957 -i0.00179*0.957 - i0.00358*0.29.$$