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If for complex numbers `z_1 and z_2, arg(z_1) -arg(z_2)=0` then `|z_1-z_2|` is equal toA. `|z_(1)|+|z_(2)|`B. `|z_(1)| - |z_(2)|`C. `||z_(1)|-|z_(2)||`D. 0 |
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Answer» Correct Answer - C We have `|z_(1)-z_(2)|^(2)=|z_(1)|^(2)+|z_(2)|^(2)-2|z_(1)||z_(2)|cos(theta_(1)-theta_(2))` where `theta_(1)=arg(z_(1))and theta_(2)=arg(z_(2))`. Given, `arg(z_(1))-arg(z_(2))=0` `rArr|z_(1)-z_(2)|^(2)=|z_(1)|^(2)+|z_(2)|^(2)-2|z_(1)||z_(2)|` `=(|z_(1)|-|z_(2)|)^(2)` `rArr|z_(1)-z_(2)|=||z_(1)|-|z_(2)||` |
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