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`Z_1!=Z_2`are two points in an Argand plane. If `a|Z_1|=b|Z_2|,`then prove that `(a Z_1-b Z_2)/(a Z_1+b Z_2)`is purely imaginary. |
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Answer» Let `Z_(1) = r_(1) e^(itheta),Z_(2) =r_(2)e^(i(theta + alpha))` Given that `ar_(1) = br_(2)` `therefore Z = (aZ_(1) - bZ_(2))/(aZ_(1) + bZ_(2))=(e^(itheta)-e^(i(theta + alpha)))/(e^(itheta) + e^(itheta+alpha))` `= (1-e^(ialpha))/(1+e^(ialpha))" "("Dividing Nr. and Dr. by" e^(itheta))` `(e^(-ialpha//2)-e^(ialpha//2))/(e^(-ialpha//2) + e^(ialpha//2))" "("Dividing Nr. and Dr.by " e^(ialpha//2))` `= (-2i sin.(alpha)/(2))/(2cos.(alpha)/(2))` `=-i tan .(alpha)/(2)` Hence, Z is purely imaginary. |
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