Verified Solution • 2025–26 Session moderate

QJEEAdv13Ma6376: JEE Advanced 2013 Mathematics - Class 11 JEEAdv13Ma6376 Solution: Let , where and .

Let S=S1∩S2∩S3S = {S_1} \cap {S_2} \cap {S_3} , where {S_1} = \left\{ {z \in C:\left| z \right| < 4} \right\},{S_2} = \left\{ {z \in C:{\mathop{\rm Im}\nolimits} \left[ {{{z - 1 + \sqrt 3 i} \over {1 - \sqrt 3 i}}} \right] > 0} \right\} and {S_3} = \left\{ {z \in C:{\mathop{\rm Re}\nolimits} z > 0} \right\}\, .

 min⁡z∈S∣1−3i−z∣=\,\mathop {\min }\limits_{z \in S} \left| {1 - 3i - z} \right| =

Multiple Choice Options

  • A 2−32{{2 - \sqrt 3 } \over 2}
  • B 2+32{{2 + \sqrt 3 } \over 2}
  • C 3−32{{3 - \sqrt 3 } \over 2}
  • D 3+32{{3 + \sqrt 3 } \over 2}

Step-by-Step Verified Solution

We know ∣1−3i−z∣=∣z−(1−3i)∣|1-3 i-z|=|z-(1-3 i)| implies distance of zz from (1,−3)(1,-3) .

Let P=(1,−3)P=(1,-3)

JEE Advanced 2013 Paper 2 Offline Mathematics - Complex Numbers Question 21 English Explanation

min⁡z∈S∣1−3i−z∣=PN\min _{z \in \mathrm{S}}|1-3 i-z|=\mathrm{PN}

amp;⇒min⁡z∈S∣1−3i−z∣=∣3⋅1−3∣(3)2+12amp;⇒min⁡z∈S∣1−3i−z∣=3−32\begin{aligned}& \Rightarrow \min _{z \in S}|1-3 i-z|=\frac{|\sqrt{3} \cdot 1-3|}{\sqrt{(\sqrt{3})^2+1^2}} \\& \Rightarrow \min _{z \in S}|1-3 i-z|=\frac{3-\sqrt{3}}{2}\end{aligned}

Hints:

(i) ∣z1−z2∣\left|z_1-z_2\right| implies distance between z1z_1 and z2z_2

(ii) The minimum distance of a point (x1,y1)\left(x_1, y_1\right) from the line ax+by+c=0a x+b y+c=0 is ∣ax1+by1+c∣a2+b2\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}} .

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