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Worked solution

Real parts, imaginary parts, moduli and arguments

Let z=2+i3−iz=\dfrac{2+i}{3-i}. Without using a calculator, find the real and imaginary parts of

(a) 1+z1−z\dfrac{1+z}{1-z},

(b) z+1zz+\dfrac1z.

Hence, find the modulus and argument of 1+z1−z\dfrac{1+z}{1-z} and z+1zz+\dfrac1z.

CJC · Discussion · Question 11

Answer

(a) Real 1, imaginary 2; modulus 5\sqrt5, argument tan⁡−12\tan^{-1}2. (b) Real 3/23/2, imaginary −1/2-1/2; modulus 5/2\sqrt{5/2}, argument −tan⁡−1(1/3)-\tan^{-1}(1/3).

(a)

1+z1−z=3−i+2+i3−i−2−i=51−2i\dfrac{1+z}{1-z}=\dfrac{3-i+2+i}{3-i-2-i}=\dfrac5{1-2i}. Rationalise to obtain 5(1+2i)1−4i2=1+2i\dfrac{5(1+2i)}{1-4i^2}=1+2i. The real part is 1 and the imaginary part is 2.

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(b)

z+1z=2+i3−i+3−i2+i=(2+i)2+(3−i)2(3−i)(2+i)=11−2i7+iz+\dfrac1z=\dfrac{2+i}{3-i}+\dfrac{3-i}{2+i}=\dfrac{(2+i)^2+(3-i)^2}{(3-i)(2+i)}=\dfrac{11-2i}{7+i}. Multiply by (7−i)/(7−i)(7-i)/(7-i) to get 77−11i−14i+2i249−i2=75−25i50=32−12i\dfrac{77-11i-14i+2i^2}{49-i^2}=\dfrac{75-25i}{50}=\dfrac32-\dfrac12i. The real part is 3/2 and the imaginary part is −1/2.

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Modulus and argument of (a)

The modulus is 12+22=5\sqrt{1^2+2^2}=\sqrt5. The first-quadrant argument is tan⁡−12\tan^{-1}2. The source also gives 1.10714 rad ≈ 1.11 rad (3 s.f.). The exact form answers the no-calculator request.

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Modulus and argument of (b)

The modulus is (3/2)2+(1/2)2=5/2\sqrt{(3/2)^2+(1/2)^2}=\sqrt{5/2}. The point is in the fourth quadrant, so its principal argument is −tan⁡−1(1/3)≈−0.32175-\tan^{-1}(1/3)\approx-0.32175 rad, or −0.322 rad (3 s.f.). The source prints the correct negative arctangent expression but drops the minus sign in its decimal lines; those positive decimals are a source error.

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Common mistakes

A fourth-quadrant number with positive real part and negative imaginary part has a negative principal argument.

Solution notes

Source discrepancy on printed/PDF p.16: arg(z+1/z)=−tan⁻¹((1/2)/(3/2)) is followed by “=0.32175 rad” and “≈0.322 rad (to 3 s.f.)”. Both printed decimal values omit the negative sign. The final student answer uses the independently checked negative principal argument. Exact argument forms are retained because the prompt says without a calculator.

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