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TS EAMCET · Maths · Complex Number

\(\text { If } z_1=x_1+i y_1, z_2=x_2+i y_2, z_3=x_1+\frac{i x_2}{2}\) \(z_4=2 y_1+i y_2\) are complex numbers such that \(\left|z_1\right|=1,\left|z_2\right|=2\) and \(\operatorname{Re}\left(z_1 z_2\right)=0\), then

  1. A \(\left|z_3\right|=1,\left|z_4\right|=2, \operatorname{Im}\left(z_3 z_4\right)=0\)
  2. B \(\left|z_3\right|=2,\left|z_4\right|=1, \operatorname{Re}\left(z_3 z_4\right)=0\)
  3. C \(\left|z_3\right|=1,\left|z_4\right|=2, \operatorname{Re}\left(z_3 z_4\right)=0\)
  4. D \(\left|z_3\right|=2,\left|z_4\right|=1, \operatorname{Re}\left(z_1 z_3\right)=\operatorname{Im}\left(z_2 z_4\right)=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left|z_3\right|=1,\left|z_4\right|=2, \operatorname{Re}\left(z_3 z_4\right)=0\)

Step-by-step Solution

Detailed explanation

We have, \(z_1=x_1+i y_1\) \(z_2=x_2+i y_2\) \(z_3=x_1+\frac{i x_2}{2}\) \(z_4=2 y_1+i y_2\) \(\left|z_1\right|=1,\left|z_2\right|=2\) and \(\operatorname{Re}\left(z_1 z_2\right)=0\) Let \(z_1=e^{i \alpha}\) and \(z_2=2 e^{i \beta}\) \(z_1 z_2=2 e^{i(\alpha+\beta)}\)…
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