For any two complex numbers z₁, z₂ and any real numbers a, b, |az₁ – bz₂|² + |bz₁ + az₂|² = ____________.

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Solution

For complex z₁ and z₂and real numbers a and b
${|a z_{1} - b z_{2}|}^{2} = (a z_{1} - b z_{2}) (\bar{a z_{1} - b z_{2}}) = (a z_{1} - b z_{2}) (a {\bar{z}}_{1} - b {\bar{z}}_{2}) = a^{2} z_{1} {\bar{z}}_{1} - a b z_{1} {\bar{z}}_{2} - a b z_{2} {\bar{z}}_{1} + b^{2} z_{2} {\bar{z}}_{2} {|a z_{1} - b z_{2}|}^{2} = a^{2} {|z_{1}|}^{2} - a b z_{1} {\bar{z}}_{2} - a b {\bar{z}}_{1} z_{2} + b^{2} z_{2} {\bar{z}}_{2} . . . (1) and {|a z_{1} + a z_{2}|}^{2} = (b z_{1} + a z_{2}) (\bar{b z_{1} + a z_{2}}) = (b z_{1} + a z_{2}) (b {\bar{z}}_{1} + a {\bar{z}}_{2}) = b^{2} z_{1} {\bar{z}}_{1} + a b z_{1} {\bar{z}}_{2} + a b z_{2} {\bar{z}}_{1} + a^{2} z_{2} {\bar{z}}_{2} = b^{2} {|z_{1}|}^{2} + a b z_{1} {\bar{z}}_{2} + a b {\bar{z}}_{1} z_{2} + a^{2} {|z_{2}|}^{2} . . . (2) ∴ from (1) and (2), {|a z_{1} - b z_{2}|}^{2} + {(b z_{1} + a z_{2})}^{2} = (a^{2} + b^{2}) ({|z_{1}|}^{2}) + (a^{2} + b^{2}) {|z_{2}|}^{2} = (a^{2} + b^{2}) ({|z_{1}|}^{2} + {|z_{2}|}^{2})$

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