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Mathematics 19 Online
OpenStudy (maheshmeghwal9):

how can we prove that:--

OpenStudy (experimentx):

prove what??

OpenStudy (maheshmeghwal9):

\[z*z'=|z|^2\] where z= complex no.

OpenStudy (anonymous):

Take z = x+ iy Then z' = x-iy Also we know that |z| = sqrt(x^2 + y^2) Substitute and solve

OpenStudy (experimentx):

let \(z = a + ib \) then \(z' = a - ib \) multiply them ,,, you will get a^2 + b^2 ... which is |z|^2

OpenStudy (anonymous):

You will get LHS = RHS

OpenStudy (maheshmeghwal9):

k! but is there no short method?

OpenStudy (maheshmeghwal9):

k!k! i have understood it. thanx to both of u:)

OpenStudy (maheshmeghwal9):

hope i could give u both medals

OpenStudy (anonymous):

yw :)

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