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. $z = i, \pm \frac{\sqrt{3}}{2} - \frac{i}{2}$

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

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