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Differential Equations 8 Online
OpenStudy (anonymous):

COMPLEX VARIABLE

OpenStudy (anonymous):

OpenStudy (anonymous):

what does that "*" represent?

OpenStudy (anonymous):

* is equal to \(\times \)

OpenStudy (anonymous):

a cross-product?

OpenStudy (anonymous):

no.., it just times

OpenStudy (anonymous):

@gerryliyana solve for RHS first by using the values of dimensional vectors given in question

OpenStudy (anonymous):

\[\vec{a}\times\vec{b}=\left|\begin{matrix} \hat{x}&\hat{y}&\hat{z}\\ u&v&0\\ x&y&0 \end{matrix}\right| =\hat{z}(uy-vx)\\ \vec{a}\cdot\vec{b}=ux+vy\\ ab = (u+iv)(x+iy)=ux+i(vx)+i(uy)+(i^2)(xy) \]

OpenStudy (anonymous):

are you sure it's right? Because: a.b=ux+vy iz(axb)=i(uy-xv) and a*b=ux-vy+i(uy+vx) so it is not equal: ux-vy+i(uy+vx) =/= ux+vy+i(uy-xv)

OpenStudy (anonymous):

\[ab=(uy-vx)+i(vx+uy)\]

OpenStudy (anonymous):

its same with my answer @myko

OpenStudy (anonymous):

yeah i think so

OpenStudy (anonymous):

\[ab=\hat{z}\cdot(\vec{a}\times\vec{b})+i(\vec{a}\cdot\vec{b})\]

OpenStudy (anonymous):

vx+uy =/= a.b @electrokid

OpenStudy (anonymous):

so what the right answer??

OpenStudy (anonymous):

oh yea.. my bad..

OpenStudy (anonymous):

that "*" in the question must be a complex conjugate

OpenStudy (anonymous):

then it seems you might have it right.

OpenStudy (anonymous):

good idea!! oh my bad...,

OpenStudy (anonymous):

yeah,"*" means conjugate

OpenStudy (anonymous):

always make sure that people are talking the same math language

OpenStudy (anonymous):

thank you @electrokid

OpenStudy (anonymous):

@electrokid if "*" is complex conjugate then it will be rightly solved

OpenStudy (anonymous):

i got it now.., thank you guys :)

OpenStudy (anonymous):

by the way.., if all of you have a facebook., add me on facebook http://www.facebook.com/gerry.resmiliyana thank you so much!

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