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Complex Numbers question

2023 · 11 Apr · Shift 1 · Q28
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Complex Numbers question

2023 · 11 Apr · Shift 1 · Q28

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let w1w_{1}w1​ be the point obtained by the rotation of z1=5+4iz_{1}=5+4 iz1​=5+4i about the origin through a right angle in the anticlockwise direction, and w2w_{2}w2​ be the point obtained by the rotation of z2=3+5iz_{2}=3+5 iz2​=3+5i about the origin through a right angle in the clockwise direction. Then the principal argument of w1−w2w_{1}-w_{2}w1​−w2​ is equal to :
  1. A
    −π+tan⁡−189-\pi+\tan ^{-1} \frac{8}{9}−π+tan−198​
  2. B
    −π+tan⁡−1335-\pi+\tan ^{-1} \frac{33}{5}−π+tan−1533​
  3. C
    π−tan⁡−189\pi-\tan ^{-1} \frac{8}{9}π−tan−198​
  4. D
    π−tan⁡−1335\pi-\tan ^{-1} \frac{33}{5}π−tan−1533​
View written solutionFree

Correct answer: C

  1. Use rotation in the complex plane

A rotation about the origin by:

  • 90∘90^\circ90∘ anticlockwise corresponds to multiplication by iii.
  • 90∘90^\circ90∘ clockwise corresponds to multiplication by −i-i−i.
  1. Find w1w_1w1​

Given z1=5+4iz_1=5+4iz1​=5+4i

After a right-angle anticlockwise rotation, w1=iz1=i(5+4i)=5i+4i2=5i−4=−4+5iw_1=i z_1=i(5+4i)=5i+4i^2=5i-4=-4+5iw1​=iz1​=i(5+4i)=5i+4i2=5i−4=−4+5i

So, w1=−4+5iw_1=-4+5iw1​=−4+5i

  1. Find w2w_2w2​

Given z2=3+5iz_2=3+5iz2​=3+5i

After a right-angle clockwise rotation, w2=(−i)z2=−i(3+5i)=−3i−5i2=−3i+5=5−3iw_2=(-i)z_2=-i(3+5i)=-3i-5i^2=-3i+5=5-3iw2​=(−i)z2​=−i(3+5i)=−3i−5i2=−3i+5=5−3i

So, w2=5−3iw_2=5-3iw2​=5−3i

  1. Compute w1−w2w_1-w_2w1​−w2​

w1−w2=(−4+5i)−(5−3i)=−9+8iw_1-w_2=(-4+5i)-(5-3i)=-9+8iw1​−w2​=(−4+5i)−(5−3i)=−9+8i

  1. Find the principal argument of −9+8i-9+8i−9+8i

The point (−9,8)(-9,8)(−9,8) lies in the second quadrant, so its principal argument is arg⁡(−9+8i)=π−tan⁡−1(89)\arg(-9+8i)=\pi-\tan^{-1}\left(\frac{8}{9}\right)arg(−9+8i)=π−tan−1(98​)

  1. Match with the options

This corresponds to: π−tan⁡−189\boxed{\pi-\tan^{-1}\frac{8}{9}}π−tan−198​​

So the correct option is C.

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