AP EAMCET · Maths · Complex Number
Match the items of List - I with those of List - II
\(\begin{array}{ll}
\text{List - I (Complex number)} & \text{List - II (Polar form)} \\
\text{(i)} \sqrt{3}-i & \text{(a)} 2 \operatorname{cis} \frac{\pi}{6} \\
\text{(ii)} \sqrt{3}+i & \text{(b)} 2 \operatorname{cis} \frac{5 \pi}{6} \\
\text{(iii)} -\sqrt{3}+i & \text{(c)} 2 \operatorname{cis}\left(\frac{-5 \pi}{6}\right) \\
\text{(iv)} -\sqrt{3}-i & \text{(d)} 2 \operatorname{cis}\left(-\frac{\pi}{6}\right) \\
& \text{(e)} 2 \operatorname{cis} \frac{9 \pi}{6} \\
\end{array}\)
The correct matching is
- A \(\begin{array}{cccc}
\text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } \\
\text { d } & \text { b } & \text { a } & \text { e } \\
\end{array}\) - B \(\begin{array}{cccc}
\text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } \\
\text { d } & \text { a } & \text { b } & \text { c } \\
\end{array}\) - C \(\begin{array}{cccc}
\text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } \\
\text { b } & \text { d } & \text { a } & \text { c } \\
\end{array}\) - D \(\begin{array}{cccc}
\text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } \\
\text { b } & \text { c } & \text { a } & \text { d }
\end{array}\)
Answer & Solution
Correct Answer
(B) \(\begin{array}{cccc}
\text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } \\
\text { d } & \text { a } & \text { b } & \text { c } \\
\end{array}\)
Step-by-step Solution
Detailed explanation
For all complex numbers \( z=x+iy \) in List-I, \( r=\sqrt{x^2+y^2}=\sqrt{(\pm\sqrt{3})^2+(\pm 1)^2} = \sqrt{3+1}=2 \). (i) \( \sqrt{3}-i \): Quadrant IV, \( \theta = -\frac{\pi}{6} \implies 2 \operatorname{cis} (-\frac{\pi}{6}) \equiv \text{(d)} \) (ii) \( \sqrt{3}+i \):…
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