InterviewSolution
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InterviewSolution
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If `z_1,z_2.........z_n=z`, then `argz_1+arg z_2+.......+arg z_n and arg z` differ by a |
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Answer» (i) Let `z=r(cos theta + i sin theta)`. Then, `|z| = r and arg(z) = theta`. Now, `z = r(cos theta +i sin theta)` `rArr" "z=r cos theta + i(r sin theta)` `rArr" "bar(z)=r cos theta-i(r sin theta)=r(cos theta -i sin theta)` `" "=r{cos(-theta)+ i sin(-theta)}` `rArr" "|bar(z)|=r and arg(bar(z))=-theta =-arg(z)`. Hence, `arg(bar(z))=-arg(z)`. (ii) Let `z_(1)=r_(1)(cos theta_(1)+ i sin theta_(1))and z_(2)=r_(2)(cos theta_(2)+ i sin theta_(2))`. Then, `|z_(1)|=r_(1), arg(z_(1))=theta and |z_(2)|=r_(2), arg(z_(2))=theta_(2)`. `therefore" "z_(1)z_(2)=r_(1)(cos theta_(1)+ i sin theta_(1)).r_(2)(cos theta_(2)+ i sin theta_(2))` `=r_(1)r_(2){(cos theta_(1)cos theta_(2)-sin theta_(1)sin theta_(2))+i(sin theta_(1)cos theta_(2)+cos theta_(1)sin theta_(2))}` `=r_(1)r_(2){cos(theta_(1)+theta_(2))+ i sin (theta_(1) + theta_(2))}` `rArr" "arg(z_(1)z_(2))=(theta_(1)theta_(2))=arg(z_(1))+arg(z_(2))`. REMARKS (I) Note here that `|z_(1)z_(2)|=r_(1)r_(2)=|z_(1)||z_(2)|`. (II) In general, we have `|z_(1)z_(2)....z_(n)|=|z_(1)|.|z_(2)|....|z_(n)| and arg(z_(1)z_(2)....z_(n))=arg(z_(1)) + arg(z_(2))+....+arg(z_(n))`. (iii) Let `z_(1)=r_(1)(cos theta_(1)+i sin theta_(1))and z_(2)=r_(2)(cos theta_(2)+ i sin theta_(2))`. Then, `bar(z)_(2)=bar(r_(2)cos theta_(2)+i(r_(2)sin theta_(2)))=r_(2)cos theta_(2)-i(r_(2)sin theta_(2))` `rArr" "bar(z)_(2)=r_(2){cos(-theta_(2))+ i sin (-theta_(2))}`. `therefore" "z_(1)bar(z)_(2)=r_(1)(cos theta_(1)+ i sin theta_(1)).r_(2){cos(-theta_(2))+ i sin (-theta_(2))}` `=r_(1)r_(2)(cos theta_(1)+ i sin theta_(1)){cos(-theta_(2))+ i sin(-theta_(2))}` `=r_(1)r_(2)[cos{theta_(1)+(-theta_(2))}+ i sin {theta_(1)+(-theta_(2)}]` `=r_(1)r_(2){cos(theta_(1)-theta_(2))+i sin(theta_(1)-theta_(2))}` Hence, `arg(z_(1)bar(z)_(2))=(theta_(1)-theta_(2))=arg(z_(1))-arg(z_(2))`. Let `z_(1)=r_(1)(cos theta_(1)+i sin theta_(1))and z_(2)=r_(2)(cos theta_(2)+i sin theta_(2))`. Then, `|z_(1)|=r_(1),|z_(2)|=r_(2), arg(z_(1))=theta_(1)and arg(z_(2))=theta_(2)`. `therefore" "(z_(1))/(z_(2))=(r_(1)(cos theta_(1)+i sin theta_(1)))/(r_(2)(cos theta_(2)+i sin theta_(2)))xx((cos theta_(2)-i sin theta_(2)))/((cos theta_(2)-i sin theta_(2)))` `=(r_(1))/(r_(2)).{((cos theta_(1).cos theta_(2)-sin theta_(1).sin theta_(2))+i(sin theta_(1).cos theta_(2)-cos theta_(1).sin theta_(2)))/((cos^(2) theta_(2)+sin^(2)theta_(2)))}` `=(r_(1))/(r_(2)).{cos(theta_(1)-theta_(2))+i sin(theta_(1)-theta_(2))}` `rArr" "arg((z_(1))/(z_(2)))=(theta_(1)-theta_(2))=arg(z_(1))-arg(z_(2))`. Hence, `arg((z_(1))/(z_(2)))=arg(z_(1))-arg(z_(2))`. |
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