[JEE Advanced 2008] p starts at z 0 1 2i moves 5 units horizontally away from the
$P$ starts at $z_0=1+2i$, moves $5$ units horizontally away from the origin, then $3$ units vertically away to reach $z_1$; then $\sqrt2$ units along $\hat i+\hat j$, then turns $\frac\pi2$ anticlockwise on a circle about the origin to reach $z_2$. Then $z_2=$
(a) $6+7i$
(b) $-7+6i$
(c) $7+6i$
(d) $-6+7i$
1 Answer
VAVidaara Admin
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· 2d ago
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Correct answer: (d) $-6+7i$
$z_0\to6+2i\to6+5i=z_1\to6+5i+(1+i)=7+6i$; rotating $7+6i$ by $\frac\pi2$ (multiply by $i$) gives $-6+7i$.
JEE Advanced 2008 · Complex Numbers — verified solution by the Vidaara Team.
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