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anonymous
 3 years ago
COMPLEX VARIABLE
anonymous
 3 years ago
COMPLEX VARIABLE

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what does that "*" represent?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0* is equal to \(\times \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@gerryliyana solve for RHS first by using the values of dimensional vectors given in question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\vec{a}\times\vec{b}=\left\begin{matrix} \hat{x}&\hat{y}&\hat{z}\\ u&v&0\\ x&y&0 \end{matrix}\right =\hat{z}(uyvx)\\ \vec{a}\cdot\vec{b}=ux+vy\\ ab = (u+iv)(x+iy)=ux+i(vx)+i(uy)+(i^2)(xy) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you sure it's right? Because: a.b=ux+vy iz(axb)=i(uyxv) and a*b=uxvy+i(uy+vx) so it is not equal: uxvy+i(uy+vx) =/= ux+vy+i(uyxv)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ab=(uyvx)+i(vx+uy)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its same with my answer @myko

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ab=\hat{z}\cdot(\vec{a}\times\vec{b})+i(\vec{a}\cdot\vec{b})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0vx+uy =/= a.b @electrokid

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so what the right answer??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that "*" in the question must be a complex conjugate

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then it seems you might have it right.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0good idea!! oh my bad...,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah,"*" means conjugate

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0always make sure that people are talking the same math language

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you @electrokid

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@electrokid if "*" is complex conjugate then it will be rightly solved

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got it now.., thank you guys :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by 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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