1. Tardigrade
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  3. Mathematics
  4. If z1=6+i,z2=4-3i and z be a complex number such that arg ( (z-z1/z2-z) )=(π /2), then z satisfies

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Q. If $ {{z}_{1}}=6+i,{{z}_{2}}=4-3i $ and z be a complex number such that arg $ \left( \frac{z-{{z}_{1}}}{{{z}_{2}}-z} \right)=\frac{\pi }{2}, $ then z satisfies

JamiaJamia 2013

Solution:

$ \because $ $ \arg \left( \frac{z-{{z}_{1}}}{{{z}_{2}}-z} \right)=\frac{\pi }{2} $ $ \Rightarrow $ $ \operatorname{Re}\left( \frac{z-{{z}_{1}}}{{{z}_{2}}-z} \right)=0 $ $ \Rightarrow $ $ \frac{z-{{z}_{1}}}{{{z}_{2}}-z}+\frac{\overline{z}-{{\overline{z}}_{1}}}{{{\overline{z}}_{2}}-\overline{z}}=0 $ $ \Rightarrow $ $ (z-{{z}_{1}})({{\overline{z}}_{2}}-\overline{z})+({{z}_{2}}+-z)(\overline{z}-{{\overline{z}}_{1}})=0 $ $ \Rightarrow $ $ z({{\overline{z}}_{1}}+{{\overline{z}}_{2}})+\overline{z}({{z}_{1}}+{{z}_{2}})2z\overline{z} $ $ -({{z}_{1}}{{\overline{z}}_{2}}+{{z}_{2}}\overline{{{z}_{1}}})=0 $ $ \Rightarrow $ $ z\overline{z}-(5+i)z+(5-i)\overline{z}+21=0 $ $ \therefore $ $ |z-(5-i)|=\sqrt{5} $