If z = a+bi then z*=a-bi, where i²=-1; a , b Real; z Complex.
z* is said to be the conjugate of z. Then zz* = a²+b² = |z|², and |z| is called the modulus (i.e. magnitude, or size).
Rules for Conjugates:
(x±y)* = x*±y*; (xy)*=x*y*; (s/t)* = s*/t*; (a zn)* = a (z*)n. for real a, complex z, integer n.
Rules for Modulus:
|xy|=|x||y|; |x/y| = |x|/|y|; |x+y|≤ |x|+|y|
If f(z) is a polynomial with real coefficients, then f(z*) = (f(z))*.
If u=a+bi is a non-real root of f(z)=0 then u*=a-bi is also a root, so the complex roots of any polynomial with real coefficients occur in conjugate pairs.
1 Establish whether the arg (-4-3i) is 3.785 or -2.498
2 Identify the modulus and argument of these: 20+21i; 48-28i; -20-30i, -37+77i
3 If z=3+I and w=5-2i then find modulus and argument of z+w and z-w
4 Using z & w from Q3, find the modulus and argument of
(i)z/w (ii)w/z, (iii)z2w3 (iv)(iz)2(iw)3(v)(iwz)2(vi)(w+iz)2.
Complex numbers can be represented on an Argand diagram, where x-axis is real and y-axis is imaginary. Modulus is then the distance of the point from the origin and argument is the angle measured anti-clockwise from the positive x-axis. To say that
argument = tan-1(b/a) is ambiguous; it is better to use the modulus-argument form where z is written as r (cosø+ isinø) and define r = |z| > 0 and ø = arg z, -π < ø ≤ π. The convention is that the principal argument is the value of tan-1(b/a) in the region -π < ø ≤ π. The convention goes further, that arg z lies in this region while Arg z is the infinite set of arguments. An alternative name for argument is amplitude. The expression (cosø+ isinø) is sometimes written as cisø.
Complex numbers on an Argand diagram can be viewed as position vectors and therefore can, by translation, be drawn as free vectors. Thus the point 5+3i could be considered as the point, as the position vector and even as the displacement vector.
5 Find the modulus and argument of:
(i) cis 5π/6 (ii) cis (-π/6) (iii) cis (π/3) (iv) cis (-π/3) (v) (cis 5π/6)²
6 Find the r (cosø+ isinø) form of:
(i) 3+√3i (ii )8+15i (iii) 24-7i (iv) -15- 20i (v) (√7+√3i)²
7 Find, and then sketch on an Argand diagram, the five roots of z⁵=1. Write each root down in both r (cosø+ isinø) form and in a+bi form.
8 Show that, by writing zj in modulus argument form, zj = rjcisøj, that z₁ z₂ = r₁r₂ cis(ø₁+ø₂). Show similarly that z₁ / z₂ = (r₁/ r₂) cis(ø₁-ø₂).
Conclude that Arg (z₁ z₂) = arg z₁ + arg z₂ and Arg (z₁ / z₂) = arg z₁ - arg z₂
9 Try to repeat Q4 without expressing them in the form a+bi first.
10 Show that zⁿ = rⁿ cis(nø). Hence find the square roots of 4√3 – 4i
There is a class of question on complex numbers based on locus (the concept of place, plural, loci). |z| < 2 is the interior of a circle radius 2, |z – z₁| < 2 the same but with centre z₁. Be clear, though, that arg (z – z₁) and arg (z₁ - z) have opposite directions and will differ by π. The argument of a line or part-line not based on the origin is the angle it makes with the (horizontal) positive real direction. Assume that, for theses next few questions, z=x+iy, so z on the Argand diagram is equivalent to (x,y) on a Cartesian sketch.
11 Sketch |z-5| = 4 and write the Cartesian equation for the equivalent point (x,y).
12 Sketch |z – 4 + i| = |z + 3 - 4i| and write the Cartesian equation.
13 Sketch arg(z – 3) – arg (z+3) = π/3 and write the Cartesian equivalent.
14 Convert the part line from (4,-3) that runs at 135º to the x-axis into a statement of locus in complex form.
15 Convert x²+4x+y²-6y+13=0 to a statement of locus in complex form.
16 Take the line ABCD that joins A(2,7) to D(5,-2), divide it into equal thirds (B&C) and define all points P (x,y) on the line perpendicular to ABCD passing through B in a statement using the language of complex numbers.
17 Draw a circle radius 13 centred on S at (0,5). Choose a general point P on the circle above the x-axis and draw the chords to the points Q & R where y=0 and draw QRS. Use that the angle QSR is twice angle QPR (= ø, if convenient) to describe the locus of P.
More to come, if prompted by a user.
18. U6 A-level, I think. Use e^iπ ≡ e iπ = cis θ to find the value of i^i ≡ i i , and hence show it is real.
10 2 cis (-π/12) ??
18 0.20788 to 5 d.p.
