TS EAMCET · Maths · Complex Number
The area (in sq units) of the triangle whose vertices are the points represented by the complex numbers \(0, z\), \(z e^{i \alpha}(0 < \alpha < \pi)\) is
- A \(\frac{1}{2}|z|^2\)
- B \(\frac{1}{2}|z|^2 \sin \alpha\)
- C \(\frac{1}{2}|z|^2 \sin \alpha \cos \alpha\)
- D \(\frac{1}{2}|z|^2 \cos \alpha\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}|z|^2 \sin \alpha\)
Step-by-step Solution
Detailed explanation
Let \(z=x+i y\) Since, vertices of triangle are \(0, z=(x+i y)\) and \[ \begin{aligned} z e^{i \alpha} & =(x+i y)(\cos \alpha+i \sin \alpha) \\ & =(x \cos \alpha-y \sin \alpha)+i(y \cos \alpha+x \sin \alpha) \end{aligned} \] \(\therefore\) Area of triangle…
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