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If `z and w` are two complex number such that `|zw|=1 and arg (z) – arg (w) = pi/2,` then show that `overline zw = -i.`A. `bar z w =i`B. `zbarw=(1-i)/sqrt(2)`C. `bar z w=i`D. `z bar w = (-1 +i)/sqrt(2)` |
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Answer» Correct Answer - A It is given that there, are two complex numbers z and w sunc that `|z w|=1 and arg(w) = pi//2` `therefore |z||w|=1 [ because |z_1 z_2|= | z_1||z_2|]` and `arg(z) = pi/2 + arg (w)` Let `|z| =r , then | w|= 1/r` and let arg (w) `= theta " then arg(z) pi/2 + theta ` So we can assume `z= re^(i pi//2+ theta)` `[ because ` if z= x+ iy is a complex number then it can be written as z `=re^(i theta) where r= |z| and theta = arg (z) ]` and `w =1/re^(it theta)` Now ` barz.w = re^(-i(pi//2theta)). 1/re^(i theta)` `=e^(i(-pi//2- theta+ theta))=e^(-i(pi//2))-i " " [ because e^(-i theta) = cos theta - i sin theta]` and `z bar w = re^(i(pi//2 + theta)).1/r e^(- i theta)` `= e^(i(pi//2+ theta - theta))=e^(i(pi//2))=i` |
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