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Complex numbers and the Argand plane

Practice section 5 of 9

Questions 25–30 in this collection

Practice questions

Worked example · 1 of 6

Sketch complex sums and prove a square

The complex numbers zz and ww are given by z=3+iz=\sqrt3+i and w=−1+i3w=-1+i\sqrt3.

Sketch an Argand diagram, with origin OO, showing the points ZZ, WW and PP representing zz, ww and z+wz+w respectively. Show that OWPZOWPZ is a square.

2010/MJC/I/11(b)(part)

CJC · Example 16

Solution

Worked example · 2 of 6

Apply the nine modulus and argument rules

The appendix works with z=1+iz=1+i and, where two numbers are needed, z1=1+iz_1=1+i and z2=3+iz_2=\sqrt3+i. Follow its nine examples: (a) zz∗zz^*; (b) ∣z∗∣\lvert z^*\rvert; (c) arg⁡z∗\arg z^*; (d) ∣z1z2∣\lvert z_1z_2\rvert; (e) ∣z4∣\lvert z^4\rvert; (f) ∣z1/z2∣\lvert z_1/z_2\rvert; (g) arg⁡(z1z2)\arg(z_1z_2); (h) arg⁡(z4)\arg(z^4); (i) arg⁡(z1/z2)\arg(z_1/z_2).

CJC · Complex Numbers · Appendix C worked examples

Solution

Discussion practice · 3 of 6

Complex roots without a graphing calculator

A graphic calculator is not to be used in answering this question.

Solve the equation z2=3−4iz^2=3-4i, giving your answers in the form a+iba+ib. Hence, by expressing 13i−14\frac13i-\frac14 in the form k(3−4i)k(3-4i), where kk is a constant, find the roots of the equation

w2=13i−14w^2=\frac13i-\frac14.

CJC · Discussion · Question 1

Solution

Discussion practice · 4 of 6

Real and imaginary cases with a conjugate

The complex number ww is such that kw2+kww∗+iw−iw∗−1=0kw^2+kww^*+iw-iw^*-1=0, where w∗w^* is the complex conjugate of ww and kk is a real and non-zero constant.

(i) For w=a+biw=a+bi where aa and bb are real numbers, obtain an expression for bb in terms of aa and kk. Explain why ww is either purely real or purely imaginary.

(ii) Using your result in part (i), or otherwise, find the real roots of the equation 2w2+2ww∗+iw−iw∗−1=02w^2+2ww^*+iw-iw^*-1=0.

[2016/DHS/I/2]

CJC · Discussion · Question 2

Solution

Discussion practice · 6 of 6

Simultaneous equations over complex numbers

Solve the following simultaneous equations:

(a) 4iz−3w=1+5i4iz-3w=1+5i and 2z+(1+i)w=2+6i2z+(1+i)w=2+6i,

(b) iz+2w=1iz+2w=1 and 4z+(3−i)w∗=−64z+(3-i)w^*=-6,

giving zz and ww in the form a+bia+bi where aa and bb are real.

CJC · Discussion · Question 4

Solution

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