Complex Numbers 2 Question 32
33.
For any two complex numbers $z _1, z _2$ and any real numbers $a$ and $b,\left|a z _1-b z _2\right|^{2}+\left|b z _1+a z _2\right|^{2}=\ldots$.
(1988, 2M)
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Answer:
Correct Answer: 33. $(a^2+b^2)(|z_1|^2)(|z_2|^2)$
Solution:
- $\left|a z _1-b z _2\right|^{2}+\left|b z _1+a z _2\right|^{2}$
$ \begin{aligned} & =\left[a^{2}\left|z _1\right|^{2}+b^{2}\left|z _2\right|^{2}-2 a b \operatorname{Re}\left(z _1 \bar{z} _2\right)\right] \\ & \quad+\left[b^{2}\left|z _1\right|^{2}+a^{2}\left|z _2\right|^{2}+2 a b \operatorname{Re}\left(z _1 \bar{z} _2\right)\right] \\ & =\left(a^{2}+b^{2}\right)\left(\left|z _1\right|^{2}+\left|z _2\right|^{2}\right) \end{aligned} $
