A) \[\left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|\]
B) \[\left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|\]
C) \[\left\| {{z}_{1}}-{{z}_{2}} \right\|\]
D) 0
Correct Answer: C
Solution :
| [c] We have, |
| \[{{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\cos ({{\theta }_{1}}-{{\theta }_{2}})\] |
| Where \[{{\theta }_{1}}\]= are \[({{z}_{1}})\]and \[{{\theta }_{2}}\]=arg \[({{z}_{2}})\].given, |
| Arg \[({{z}_{1}}-{{z}_{2}})\]=0. |
| \[\Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\] \[={{(\left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|)}^{2}}\] \[\Rightarrow \left| {{z}_{1}}-{{z}_{2}} \right|\left. = \right|\left| {{z}_{1}}- \right|\left. \left. {{z}_{2}} \right| \right|\] |
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