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MCQEasyJEE 2023Modulus & Argument

JEE Mathematics 2023 Question with Solution

Let w1w_1 be the point obtained by the rotation of z1=5+4iz_1 = 5 + 4i about the origin through a right angle in the anticlockwise direction, and w2w_2 be the point obtained by the rotation of z2=3+5iz_2 = 3 + 5i about the origin through a right angle in the clockwise direction. Then the principal argument of w1−w2w_1 - w_2 is equal to:

  • A

    π−tan⁡−1(89)\pi - \tan^{-1}\left( \frac{8}{9} \right)

  • B

    π−tan⁡−1(489)\pi - \tan^{-1}\left( \frac{48}{9} \right)

  • C

    π−tan⁡−1(335)\pi - \tan^{-1}\left( \frac{33}{5} \right)

  • D

    π−tan⁡−1(335)\pi - \tan^{-1}\left( \frac{33}{5} \right)

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: z1=5+4iz_1 = 5 + 4i and z2=3+5iz_2 = 3 + 5i.

Find: The principal argument of w1−w2w_1 - w_2.

A rotation by 90∘90^\circ anticlockwise corresponds to multiplication by ii, and a rotation by 90∘90^\circ clockwise corresponds to multiplication by −i-i.

w1=iz1=i(5+4i)=−4+5iw_1 = i z_1 = i(5 + 4i) = -4 + 5iw2=−iz2=−i(3+5i)=5−3iw_2 = -i z_2 = -i(3 + 5i) = 5 - 3i

Now,

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

So the complex number lies in the second quadrant. Therefore its principal argument is

arg⁡(w1−w2)=π−tan⁡−1(89)\arg(w_1 - w_2) = \pi - \tan^{-1}\left(\frac{8}{9}\right)

Therefore, the correct option is A.

The solution contains an arithmetic inconsistency in the intermediate line for w2w_2 and in the printed subtraction, but the final option selected by the solution is consistent with the correct computation above.

Using rotation rule directly

Given: z=x+yiz = x + yi.

Find: The rotated forms of z1z_1 and z2z_2, then the argument of their difference.

For a complex number:

  • rotation by 90∘90^\circ anticlockwise gives −y+xi-y + xi,
  • rotation by 90∘90^\circ clockwise gives y−xiy - xi.

Applying this to z1=5+4iz_1 = 5 + 4i,

w1=−4+5iw_1 = -4 + 5i

Applying this to z2=3+5iz_2 = 3 + 5i,

w2=5−3iw_2 = 5 - 3i

Hence,

w1−w2=(−4+5i)−(5−3i)w_1 - w_2 = (-4 + 5i) - (5 - 3i)=−9+8i= -9 + 8i

Its real part is negative and imaginary part is positive, so it lies in quadrant II. Therefore,

Principal argument=π−tan⁡−1(89)\text{Principal argument} = \pi - \tan^{-1}\left(\frac{8}{9}\right)

Therefore, the principal argument is π−tan⁡−1(89)\pi - \tan^{-1}\left( \frac{8}{9} \right).

Common mistakes

  • Using the wrong rotation factor. Anticlockwise 90∘90^\circ means multiply by ii, while clockwise 90∘90^\circ means multiply by −i-i. Reversing these changes both rotated points and gives the wrong argument.

  • Making a sign error in −i(3+5i)-i(3 + 5i). Since −i⋅5i=−5i2=5-i \cdot 5i = -5i^2 = 5, the correct value is 5−3i5 - 3i, not 5+3i5 + 3i. Always use i2=−1i^2 = -1 carefully.

  • Finding tan⁡−1(yx)\tan^{-1}\left(\frac{y}{x}\right) without adjusting for the quadrant. For −9+8i-9 + 8i, the point lies in quadrant II, so the principal argument is π−tan⁡−1(89)\pi - \tan^{-1}\left(\frac{8}{9}\right), not a negative acute angle.

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