Complex Numbers: Diagram-Based Questions (5)

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Tags Complex Numbers

Question

In the Argand plane, point AA corresponds to z1=3+4iz_1 = 3 + 4i and point BB corresponds to z2=1+iz_2 = 1 + i. Find the complex number representing the midpoint of ABAB, and the modulus and argument of z1−z2z_1 - z_2 (which represents the displacement from BB to AA).

Solution — Step by Step

Midpoint of ABAB in the Argand plane corresponds to:

zM=z1+z22=(3+4i)+(1+i)2=4+5i2=2+2.5iz_M = \frac{z_1 + z_2}{2} = \frac{(3+4i) + (1+i)}{2} = \frac{4 + 5i}{2} = 2 + 2.5i z1−z2=(3−1)+(4−1)i=2+3iz_1 - z_2 = (3 - 1) + (4 - 1)i = 2 + 3i ∣z1−z2∣=4+9=13|z_1 - z_2| = \sqrt{4 + 9} = \sqrt{13}

This is the distance from BB to AA in the plane.

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

Both real and imaginary parts are positive, so the argument is in the first quadrant.

tan⁡−1(1.5)≈56.31∘≈0.983 rad\tan^{-1}(1.5) \approx 56.31^\circ \approx 0.983 \, \text{rad}

Final answer: midpoint =2+2.5i= 2 + 2.5i; ∣z1−z2∣=13|z_1 - z_2| = \sqrt{13}; arg⁡=tan⁡−1(3/2)≈56.31∘\arg = \tan^{-1}(3/2) \approx 56.31^\circ.

Why This Works

Complex numbers and 2D vectors are isomorphic — addition, midpoint, and displacement work identically in both. The Argand plane lets you visualise complex arithmetic geometrically.

The modulus is the distance from origin (or, for differences, distance between two points). The argument is the angle the vector makes with the positive real axis, measured anticlockwise.

Alternative Method

Convert to polar form first: z1−z2=2+3iz_1 - z_2 = 2 + 3i. Polar: r=13r = \sqrt{13}, θ=tan⁡−1(3/2)\theta = \tan^{-1}(3/2). The polar representation directly gives modulus and argument.

When subtracting complex numbers geometrically, z1−z2z_1 - z_2 is the vector from z2z_2 to z1z_1 (point at the head). Useful for problems involving rotation about a non-origin point.

Students compute tan⁡−1\tan^{-1} without checking the quadrant of (x,y)(x, y). For z=−2+3iz = -2 + 3i, tan⁡−1(3/(−2))=tan⁡−1(−1.5)\tan^{-1}(3/(-2)) = \tan^{-1}(-1.5) which the calculator gives in the fourth quadrant. The actual argument is in the second quadrant: π−tan⁡−1(1.5)\pi - \tan^{-1}(1.5). Always check the signs of real and imaginary parts.

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