SATHEE: Complex Numbers 2 Question 32

Complex Numbers 2 Question 32

33.

For any two complex numbers $z_{1}, z_{2}$ and any real numbers $a$ and $b, {| a z_{1} - b z_{2} |}^{2} + {| b z_{1} + a z_{2} |}^{2} = \dots$ .

(1988, 2M)

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Answer:

Correct Answer: 33. $(a^{2} + b^{2}) (| z_{1} |^{2}) (| z_{2} |^{2})$

Solution:

${| a z_{1} - b z_{2} |}^{2} + {| b z_{1} + a z_{2} |}^{2}$

$\begin{aligned} = [a^{2} {| z_{1} |}^{2} + b^{2} {| z_{2} |}^{2} - 2 a b Re (z_{1} {\bar{z}}_{2})] \\ + [b^{2} {| z_{1} |}^{2} + a^{2} {| z_{2} |}^{2} + 2 a b Re (z_{1} {\bar{z}}_{2})] \\ = (a^{2} + b^{2}) ({| z_{1} |}^{2} + {| z_{2} |}^{2}) \end{aligned}$

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Complex Numbers 2 Question 32

33.

Answer:

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