JEE Advanced · Mathematics · 3. Complex Numbers
A particle \(P\) starts from the point \(z_0=1+2 i\), where \(i=\sqrt{-1}\). It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point \(z_1\). From \(z_1\) the particle moves \(\sqrt{2}\) units in the direction of the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and then it moves through an angle \(\frac{\pi}{2}\) in anti-clockwise direction on a circle with centre at origin, to reach a point \(z_2\). The point \(z_2\) is given by
- A \(6+7 i\)
- B \(-7+6 i\)
- C \(7+6 i\)
- D \(-6+7 i\)
Answer & Solution
Correct Answer
(D) \(-6+7 i\)
Step-by-step Solution
Detailed explanation

\(z_2^{\prime}=\left(6+\sqrt{2} \times \cos 45^{\circ}, 5+\sqrt{2} \sin 45^{\circ}\right)=\) \((7,6)=7+6 i\)
By rotation about \((0,0)\),
\(\frac{z_2}{z_2^{\prime}}=e^{i \frac{\pi}{2}} \)
\( \Rightarrow z_2=z_2^{\prime}\left(e^{i \frac{\pi}{2}}\right) \)
\( \Rightarrow z_2=(7+6 i)\left(\cos \frac{\pi}{-}+i \sin \frac{\pi}{-}\right)=\) \((7+6 i)(i)=-6+7 i\)
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