Verified Solution • 2025–26 Session moderate

QJEEM22Ma0cc1: JEE Main 2022 Mathematics - Class 11 JEEM22Ma0cc1 Solution: Let O be the origin and A be the point . If B is the point , ,…

Let O be the origin and A be the point z1=1+2i{z_1} = 1 + 2i . If B is the point z2{z_2} , {\mathop{\rm Re}\nolimits} ({z_2}) < 0 , such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT true?

Multiple Choice Options

  • A arg⁡z2=π−tan⁡−13\arg {z_2} = \pi - {\tan ^{ - 1}}3
  • B arg⁡(z1−2z2)=−tan⁡−143\arg ({z_1} - 2{z_2}) = - {\tan ^{ - 1}}{4 \over 3}
  • C ∣z2∣=10|{z_2}| = \sqrt {10}
  • D ∣2z1−z2∣=5|2{z_1} - {z_2}| = 5

Step-by-Step Verified Solution

JEE Main 2022 (Online) 26th July Morning Shift Mathematics - Complex Numbers Question 68 English Explanation

z2−0(1+2i)−0=∣OB∣∣OA∣eiπ4{{{z_2} - 0} \over {(1 + 2i) - 0}} = {{|OB|} \over {|OA|}}{e^{{{i\pi } \over 4}}}

⇒z21+2i=2eiπ4\Rightarrow {{{z_2}} \over {1 + 2i}} = \sqrt 2 {e^{{{i\pi } \over 4}}}

OR z2=(1+2i)(1+i){z_2} = (1 + 2i)(1 + i)

=−1+3i= - 1 + 3i

arg⁡z2=π−tan⁡−13\arg {z_2} = \pi - {\tan ^{ - 1}}3

∣z2∣=10|{z_2}| = \sqrt {10}

z1−2z2=(1+2i)+2−6i=3−4i{z_1} - 2{z_2} = (1 + 2i) + 2 - 6i = 3 - 4i

arg⁡(z1−2z2)=−tan⁡−143\arg ({z_1} - 2{z_2}) = - {\tan ^{ - 1}}{4 \over 3}

∣2z1−z2∣=∣2+4i+1−3i∣=∣3+i∣=10|2{z_1} - {z_2}| = |2 + 4i + 1 - 3i| = |3 + i| = \sqrt {10}

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