FP2 further complex numbers
- Click here for pdf booklet for Möbius transformations lessons (2016-7 version). 2017-8 version here
- Click here for exercises from the textbook, and answers.
- Click here for recent FP2 exam questions on this topic
- Click here for worked answers to all recent FP2 exam questions on this topic
WORK COVERED IN LAST TERM OF 2015-6
Lesson 1: working out w=f(z) transformations by changing the subject of the equation
Starter: draw |z|=1, |z|=4, arg(z)=π/2, arg(z)=π/4, arg(z)=0
Write the Cartesian equations
Write Cartesian equation for |z-2|=3
Textbook pages 53-58
Classwork/ homework: Ex.3H Q.1, 3, 4, 6
Lesson 2: how to picture complex functions of a complex variable, ℂ↦ℂ
Booklet pages 1 to 3
Picturing complex functions of a complex variable works well only for simple functions. Even w=z2 turns out complicated, and the w-locus overlaps itself for many z-locuses.
w=z* looks quite simple, but isn’t so much so. You can’t define dw/dz, because δw/δz=1 if δz is a small increase in z along the real axis, but something quite different, δw/δz=−1 if δz is a small increase in z along the imaginary axis.
So a lot of our work with z-planes and w-planes (pretty much all we do in FP2) is with simpler equations, i.e. Möbius transformations.
Lesson 3: the rules for Möbius transformations (when pole not on z-locus)
Booklet pages 4 to 7
Lesson 4: To get back up to speed on Möbius transformations
Do now: On the whiteboards, write down:
1. What is the name for transformations like w = (az+b)⁄(cz+d)?
2. That transformation can also be written as z ↦ (az+b)⁄(cz+d). What does the symbol ↦ mean, in words?
3. Those transformations distort some shapes. But what simple shape C do they always transform into (another example of) the same shape?
4. That is true only when you interpret some things L which usually you’d interpret as different shapes as extreme examples of that simple shape. What things?
5. How do you tell when a transformation z ↦ (az+b)⁄(cz+d) will produce an extreme shape L?
6. What are the steps for finding the image shape by the geometrical methods?
7. What are the steps for finding the image shape by the algebraic method?
8. All of these transformations are combinations of three simpler sorts of transformations. What are those three?
Practice: Activity 10, page 10 of the booklet
Lesson 5: what Möbius transformations do to regions; how to interpret “diameter ends” of the z-locus when it is a line.
If w = (z−i)/(z+i), then the z-locus |z|=2 (left plane) produces the w-locus shown in the right plane, above.
Choose a point (easy to calculate) inside, or a point outside, the circle |z|=2, and see what their images are in the w-plane. What region in the w-plane is the image of the region inside the circle |z|=2?
Activity 11, page 11 of workbook.
When the z-locus is a line, the “diameter ends” are:
1. the point on the z-line nearest to the pole
Lesson 6: complex-number equations for perpendicular bisector in ℂ, half-lines in ℂ, arcs of circles in ℂ; revisit algebraic method
Do now: Fill in this flowchart for the geometric method of calculating Möbius transformations
(If the z-locus is a line, its “diameter-ends” are ∞ and the point on the z-line nearest to the pole, which is the foot of the perpendicular line from the pole to the z-line).
Why does it make a difference to the method whether the pole is on the z-locus?
If w=(z−i)/(z+i), calculate w for these z-values:
z = ∞
z = −∞
z = ∞i
z = −∞i
So in this sort of working, there is only one infinity. Lines go to Hell in many different directions, but they all end up at the same Hell. ∞ and −∞i are the same for these purposes.
Do Activity 11.
ACTIVITY 12 from workbook: In your book, draw a diagram of the points z for which |z−1|=|z−3|.
Write down the general rule for the path of z if |z−a|=|z−b|.
ACTIVITY 13 from workbook: |z|=r is the circle with centre at the origin and radius r.
What is the path of z if |z−a−ib|=r?
If z=x+iy, use the fact that |x+iy−a−ib|=|(x−a)+i(y−b)| to write an equation in x and y for the path of z.
Equations of half-lines
Work through Activity 13 in workbook
Do Activities 14 and 15 in workbook
Homework: past-paper questions on ℂ↦ℂ
Click here for worked questions.
REVISIT ALGEBRAIC METHOD FOR MÖBIUS TRANSFORMATIONS
w = (z−1)/(z+1). The locus of z is the circle |z|=1. Find the locus of w.
This is quick and easy by the algebraic method, so try the algebraic method. Then check by the geometric method.
Check the differences between geometric method and algebraic method. Review problems from the homework.
What are ln and exp? Brainstorm what you know.
log2x is the power p to which you have to raise 2 to get x: 2p=x
So log21 =
Graph of log2x and graph of 2x
The base for logs e (approx. 2.71828) is defined by: the rate of increase of ex is equal to ex itself, for all x
Or: d/dx(ex) = ex
exp(x) is another way of writing ex. ln(x) is another way of writing logex.
log10x used to be very important for practical calculations.
We’ve glossed over something: what does e.g. e√2 even mean? Is there any power of 2 that equals 3, for example? There’s another way of defining ln and exp which avoids this problem.
Do now: draw table of integrals of tn for n=−4 to n=4
Define ln x = integral from 1 to y of t−1 dt. Sketch graph of ln x
Deduce, graphically, ln(xy) = ln x + ln y (use example ln 6 = ln 2 + ln 3)
2×2=4, so ln(2×2)=ln(4). Divide ln(4) into two areas, the integral from 1 to 2, and the integral from 2 to 4.
But the integral from 2 to 4 is the same as the integral from 1 to 2, only squashed down into half-height and stretched out into double-length.
If ln(e)=1, then ln(x)=k defined by ek=x
log10x=m defined by 10m=x
Define exp(x) as inverse function of ln. Sketch graph.
Deduce exp(x).exp(y) = exp(x+y)
Deduce exp(x) = [exp(1)]x. Define e=exp(1). Then exp(x)=ex.
10x=exp[x ln 10]
Deduce (d/dx) exp(x) = exp(x) and (d/dx) exp(Ax) = A exp(x)
LN, EXP, AND EULER’S FORMULA
Do now: Write down neatly in your book, on a new page, all the rules you know about ln x and ex
[Reminder: exp(x) is another way of writing ex]
ln x = integral from 1 to x of (1/t) dt
ln x = a ⇔ ea = x
ln xy = ln x + ln y
ln (xa) = a ln x
ln (ex) = x
(d/dx) ex = ex
ea+b = ea.eb
eln x = x
(d/dx) eAx = a eAx
If z=r cis θ what are z2 and z3? and z½?
zn = rn cis nθ
So cis n = [cis 1]n
Deduce cis x = [cis 1]x for all x
What is (cis 1)?
Define [exp(t)]i = exp (it) by (d/dt) exp (it) = i exp(it) and exp(0)=1
Then the path of exp(it) is at right angles to exp(it) for all t, so if we sketch exp(it), it goes round the unit circle at a rate of 1 radian per unit of time
Therefore exp(i) = cis 1, exp(it) = cis t [Euler’s formula], eiπ+1=0
exp(kiθ)=(cos θ + i sin θ)k=cos kθ + i sin kθ
[This is called De Moivre’s theorem]
(d/dt)cos t = − sin t; (d/dt) sin t = cos t
cos t = ½(eit+e−it); sin t = ½(eit−e−it);
Classwork and homework: From your “Further complex numbers: exercises and answers” booklet, do:
1. “Exercise on exp and ln functions”, ninth page in the booklet (with “41” on bottom right corner of photocopy)
2. “Exercise 3A”, eleventh page in the booklet (with “23” on bottom right corner of photocopy)
3. More Möbius questions if I’ve asked for them in my comments on your last homework
- “Designing a polynomial” from Underground Maths
- (1+x)−1 = 1−x+x2−x3+x4−….. (geometric progression). “Design a polynomial” so that its derivative is (1+x)−1. What is the infinite series for ln(1+x)?
Using Euler’s identity and De Moivre
We know: exp (iθ) = cis θ = cos θ + i sin θ [Euler’s identity}
exp(kiθ) = (cos θ + i sin θ)k = cos kθ + i sin kθ [De Moivre’s theorem]
(d/dθ)cos θ = − sin θ; (d/dθ) sin θ = cos θ
Do now: 1. Write the “student response” on your homework
2. What is e2iπ? What is e4iπ? What is e6iπ?
3. For homework you calculated sin θ = (eiθ − eiθ)/2i
Calculate a similar formula for cos θ
4. You know exp(i[θ+φ]) = exp iθ . exp iφ
so: cos (θ+φ) + i sin (θ+φ) = (cos θ + i sin θ).(cos φ + i sin φ)
Multiply out the right hand side to get formulas for cos (θ+φ) and sin (θ+φ) in terms of cos θ, cos φ, sin θ, and sin φ
Proof that multiplying complex numbers adds arguments
Quick check on Ex. 3B, 3C
Ex. 3D: getting expressions for cos nθ and cosnθ using De Moivre
cos 30 + i sin 3θ = (cos θ + i sin θ)3 (by De Moivre)
= cos3θ + 3i cos2θ . sin θ − 3 cos θ . sin2θ − i sin3θ (by binomial expansion using Pascal’s triangle)
Equate imaginary parts: sin 3θ = 3 cos2θ . sin θ − sin3θ
= 3(1−sin2θ)sin θ − sin3θ = 3 sin θ − 4 sin3θ.
cos4θ = [½(eiθ+e−iθ]4 (from Euler’s identity)
= (1/16) (e4iθ+4e2iθ+6+4e−2iθ+e−4iθ) (by binomial expansion using Pascal’s triangle)
= (1/16) (2 cos 4θ + 8 cos 2θ + 6) = (1/8) (cos 4θ + 4 cos 2θ + 3)
Classwork and homework: Q.2, 3, 5, 6, 7 of Ex.3D (p.36); Q.1, Q.2, Q.3 a, b, c, Q.4, of Ex.3I (p.61)
Finding roots of complex numbers
Do now: find roots of z3=8 and draw them
And roots of z3=27?
And roots of z4=16?
How many cube roots does every number have in ℂ? How many fourth roots, etc.? What do they look like?
Finding cube (etc.) roots of any complex number, textbook p.38-39
Edexcel proof of De Moivre’s theorem
De Moivre’s theorem is: (cos θ + i sin θ)n = (cos nθ + i sin nθ)
Working backwards: Edexcel proof of De Moivre’s theorem, textbook p.28-29