Subjective Type

For any two complex numbers $$z_1, z_2$$ and any real numbers a & b; $$|az_1 - bz_2|^2 + |bz_1 + az_2|^2 = k(a^2 + b^2) (|z_1|^2 + |z_2|^2)$$. What is the value of $$2k$$?

Solution

$$|az_{ 1 }-bz_{ 2 }|^{ 2 }+|bz_{ 1 }+az_{ 2 }|^{ 2 }=\left( a{ z }_{ 1 }-b{ z }_{ 2 } \right) \left( a\bar { { z }_{ 1 } } -b\bar { { z }_{ 2 } } \right) +\left( a{ z }_{ 1 }+b{ z }_{ 2 } \right) \left( a\bar { { z }_{ 1 } } +b\bar { { z }_{ 2 } } \right) $$
$$={ a }^{ 2 }{ \left| { z }_{ 1 } \right| }^{ 2 }-ab{ z }_{ 1 }\bar { { z }_{ 2 } } -ab{ z }_{ 2 }\bar { { z }_{ 1 } } +{ b }^{ 2 }{ \left| { z }_{ 2 } \right| }^{ 2 }+{ a }^{ 2 }{ \left| { z }_{ 1 } \right| }^{ 2 }+ab{ z }_{ 1 }\bar { { z }_{ 2 } } +ab{ z }_{ 2 }\bar { { z }_{ 2 } } +{ b }^{ 2 }{ \left| { z }_{ 2 } \right| }^{ 2 }$$
$$\Rightarrow |az_{ 1 }-bz_{ 2 }|^{ 2 }+|bz_{ 1 }+az_{ 2 }|^{ 2 }=(a^{ 2 }+b^{ 2 })(|z_{ 1 }|^{ 2 }+|z_{ 2 }|^{ 2 })$$

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