Complex Numbers: Application Problems (1)

easy • 2 min read
Tags Complex Numbers

Question

If z=1+2i1−iz = \tfrac{1 + 2i}{1 - i}, find the modulus and argument of zz.

Solution — Step by Step

Multiply numerator and denominator by the conjugate of the denominator, 1+i1 + i:

z=(1+2i)(1+i)(1−i)(1+i)z = \tfrac{(1 + 2i)(1 + i)}{(1 - i)(1 + i)} (1−i)(1+i)=1−i2=1−(−1)=2(1 - i)(1 + i) = 1 - i^2 = 1 - (-1) = 2 (1+2i)(1+i)=1+i+2i+2i2=1+3i−2=−1+3i(1 + 2i)(1 + i) = 1 + i + 2i + 2i^2 = 1 + 3i - 2 = -1 + 3i z=−1+3i2=−12+32iz = \tfrac{-1 + 3i}{2} = -\tfrac{1}{2} + \tfrac{3}{2}i ∣z∣=14+94=104=102|z| = \sqrt{\tfrac{1}{4} + \tfrac{9}{4}} = \sqrt{\tfrac{10}{4}} = \tfrac{\sqrt{10}}{2}

For the argument: zz has real part −1/2-1/2 (negative) and imaginary part 3/23/2 (positive), so zz lies in the second quadrant.

tan⁡θref=3/21/2=3  ⟹  θref=tan⁡−1(3)\tan\theta_{\text{ref}} = \tfrac{3/2}{1/2} = 3 \implies \theta_{\text{ref}} = \tan^{-1}(3)

In the second quadrant, arg⁡z=π−tan⁡−1(3)\arg z = \pi - \tan^{-1}(3).

Final answers: ∣z∣=102|z| = \mathbf{\tfrac{\sqrt{10}}{2}}, arg⁡z=π−tan⁡−1(3)≈1.893\arg z = \pi - \tan^{-1}(3) \approx \mathbf{1.893} rad.

Why This Works

Multiplying numerator and denominator by the conjugate kills the imaginary part of the denominator (since (a−bi)(a+bi)=a2+b2(a-bi)(a+bi) = a^2+b^2, a real number). This is the standard trick to convert any complex fraction to a+bia + bi form.

Once in standard form, modulus is a2+b2\sqrt{a^2 + b^2} (Pythagoras on the Argand diagram). The argument requires checking the quadrant — tan⁡−1(b/a)\tan^{-1}(b/a) alone gives only the principal value, which may need ±π\pm\pi correction.

Alternative Method

Use polar form. Numerator 1+2i1 + 2i: modulus 5\sqrt{5}, argument tan⁡−1(2)\tan^{-1}(2). Denominator 1−i1 - i: modulus 2\sqrt{2}, argument −π/4-\pi/4. Then:

∣z∣=52=102✓|z| = \tfrac{\sqrt{5}}{\sqrt{2}} = \tfrac{\sqrt{10}}{2} \checkmark arg⁡z=tan⁡−1(2)−(−π/4)=tan⁡−1(2)+π/4\arg z = \tan^{-1}(2) - (-\pi/4) = \tan^{-1}(2) + \pi/4

This is equivalent to π−tan⁡−1(3)\pi - \tan^{-1}(3) via the identity tan⁡−1(2)+tan⁡−1(3)=π−tan⁡−1(1)=3π/4\tan^{-1}(2) + \tan^{-1}(3) = \pi - \tan^{-1}(1) = 3\pi/4.

Common Mistake

Reporting arg⁡z=tan⁡−1(b/a)\arg z = \tan^{-1}(b/a) without checking the quadrant. For z=−1/2+(3/2)iz = -1/2 + (3/2)i, tan⁡−1(b/a)=tan⁡−1(−3)≈−1.249\tan^{-1}(b/a) = \tan^{-1}(-3) \approx -1.249 rad — but this puts the angle in quadrant IV, not II where zz actually lies. The sign of the components decides the quadrant, not the sign of the ratio.

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