Differential Equations: Lecture 1

The term differential equation was coined by Leibniz in 1676 for a relationship between the two differentials dx and dy for the two variables x and y.

A differential equation is an equation which involves an unknown function (e.g. y) and some of its derivatives. Solving a differential equation (DE) is to find a relation between the variables, say y and x, that satisfies the DE.

In today's lecture, we have learnt the followings:

1. Terminology: what is the order and degree of a differential equation?

2. How to solve the following two types of DEs?

• \[\frac{\mathrm{d} y}{\mathrm{d} x}=f(x)\]

To get the solution, we use \[y=\int f(x) dx\]

• \[\frac{\mathrm{d} y}{\mathrm{d} x}=f(y)\]

To get the solution, we use \[\int \frac{1}{f(y)} dy=x\]

3. General solutions and particular solutions

A differential equation has infinite no. of solutions. For example, \[y=x^{2}+1,y=x^{2}+2 \] are both the solutions for the DE \[\frac{\mathrm{d} y}{\mathrm{d} x}=2x\]. Indeed, \[y=x^{2}+C\] will be a solution for the DE with any constant C.

In this case, we call \[y=x^{2}+C\] a general solution.

If a condition on x and y is given, say y=1 when x=0, then we can find out a specific value of C and that is a particular solution.

After today's lecture, you can attempt Q1-Q7 in tutorial 11a.

Complex numbers: Lecture 7

In this lecture you have learnt how to draw the loci in the form of \[arg(z-a)=\theta\]

Remember the description of such locus:

Let A and P be the points in the Argand diagram representing a and z respectively,

The locus is the half-line from A (excluding A) that makes an angle theta with the positive real axis.

In addition, we need to know the following geometrical knowledge:

1. Given a point A outside a circle with centre O, find a point B on the circle such that AB is longest.

Answer: B is the point such that AB passes through O.

2. Given a point A outside a circle with centre O, find a point B on the circle such that AB is shortest.

Answer: B is the point such that the extended AB passes through O.

3. Given a point A outside a circle with centre O, find a point B on the circle such that angle BAO is maximum.

Answer: B is the point such that AB is the tangent to the circle. (There are two such point B.)

This is the last lecture on complex numbers. Now you should finish the tutorial as soon as possible.

Complex numbers: Lecture 6

We learnt 3 important things in today's lecture.
1. Geometrical interpretation of multiplication/division of complex numbers
2. Loci in the form of |z-a|=r
3. Loci in the form of |z-a|=|z-b|

1. Geometrical interpretation of multiplication/division of complex numbers

• Multiplying a complex number z by i is equivalent of rotating z anti-clockwise about the origin by pi/2
• Multiplying a complex number z by -1 is equivalent of rotating z anti-clockwise about the origin by pi
• In general, multiplying a complex number z by another complex number with modulus k and argument x is equivalent of rotating z anti-clockwise about the origin by x and scale z by a factor of k.

2. Loci in the form of |z-a|=r

• Locus of |z-a|=r means the collection of all points that satisfy the equation, i.e. all the points whose distance to complex number a is r.
• Points with fixed distance to a given point a forms a circle with center at a and radius r.
• If the equation is changed to an inequality |z-a|>r or |z-a|< r, it means all the points whose distance to a is greater or smaller than r, which is the region outside or the interior region of the circle.

3. Loci in the form of |z-a|=|z-b|

• Locus of |z-a|=|z-b| means the collection of all points that satisfy the equation, i.e. all the points whose distances to a and b are equal.
• Points with equal distances to two fixed points a and b forms the perpendicular bisector of the line ab.
• If the equation is changed to an inequality |z-a|<|z-b|,it means all the points whose distance to a is smaller than the distance to b, which is the region on one side of the perpendicular bisector that is closer to a.
• If the equation is changed to an inequality |z-a|>|z-b|, it means all the points whose distance to a is greater than the distance to b, which is the region on one side of the perpendicular bisector that is closer to b.

Remark: when you draw the loci of an inequality, take note of whether the equal sign is included. If the circle or the perpendicular bisector itself are not included, you have to use dotted line.

After today's lecture, you can do the whole tutorial 10b and Q1,2, 4, 6, 8 and 9 in tutorial 10c.

Complex Numbers: Lecture 5

In this lecture, there are two main concepts.

1. Exponential form of a complex number
2. De Moivre's Theorem

Exponential form

You have already learnt the polar form of a complex number, i.e. \[z=r(cos\theta+i sin\theta)\]

By using the Euler's Equation: \[e^{i\theta}=cos\theta+i sin\theta\]
We can change polar form to exponential form \[z=re^{i\theta}\]

The laws of exponents for real numbers still hold for complex numbers. Hence, when complex numbers are written in exponential form, it is comparatively easy to do the computation.

De Moivre's Theorem

Basically De Moivre's Theorem tells us that \[(cos\theta+i sin\theta)^{n}=cos n\theta+i sin n\theta\] for all n being rational numbers.

It is directly derived from the exponential form of complex numbers.

De Moivre's Theorem is very useful in terms of simplifying trigonometric calculation.

After today's lecture, you can do all the questions in tutorial 10b except Q13.

Complex Numbers: Lecture 4

There are mainly two learning points in this lecture.

1. How to make use of the property of modulus and argument
2. the polar form of complex numbers.


In general, the property of argument is similar to the laws of logarithms. The properties are very important. Make sure that you are familiar with them and do not get confused between the properties of modulus and argument.

For the polar form of complex number, \[z=r(sin\theta+icos\theta)\], where r is the modulus (distance from z to the origin in argand diagram) and \[\theta\] is the argument (angle between the complex number and the positive x-axis in argand diagram)

Given the polar form, you can get the cartesian form easily by calculating \[sin\theta\] and \[cos\theta\].

Given the cartesian form, if you want to get the polar form, find the modulus and the argument first.

After this lecture, you can attempt tutorial 10b Q1, Q3, Q5, Q6, Q7 and Q13

Complex Numbers: Lecture 3

In today's lecture, you learnt two important concepts of complex numbers: moduli and argument.

1. Moduli

• Modulus of a complex number z, denoted by z, represents the distance between z and origin in the argand diagram. Hence, we have

- |z| is non-negative

- |z|=0 if and only if z=0

• z1-z2 or z2-z1 represents the distance between z1 and z2 in the argand diagram.

2. Argument

• If z is represented by point P in the argand diagram, argument of z is the angle between OP and the positive real axis.
• The principal argument of z is in the interval (-pi, pi]. Take note that -pi is not included by pi is included! (why not include both?)
• To find the argument of z=x+yi

- find the basic angle using \[tan^{-1}\frac{y}{x}\]

(If you don't know what basic angle is, please revise your secondary school work.)

- determine which quadrant z lies in and use the basic angle to get the argument

3. Properties of argument and modulus

• Properties of argument is similar to the properties of logarithm. This is not by coincidence. After you learnt the exponential form of complex numbers, you will know the reason.

After today's lecture, you can attempt Q1 and Q3 in tutorial 10b.

Complex Numbers: Lecture 2

In today's lecture, there are three main concepts.

1. Complex conjugates

• If z=x+yi, then its conjugate is x-yi, denoted by z*.
• z and z* are reflections along the x-axis.
• \[ \left |z|\right =\sqrt{x^{2}+y^{2}}\], where z is called the modulus of z. (You can think of z as a vector in the argand diagram. Do you still remember the modulus of a vector?)
• \[ zz^{*}=z^{2} \]

The seven properties of the complex conjugates are very important. Please make use that you learn them by heart.

2. Division of complex numbers

• to calculate z1/z2, multiply it by z2*/z2* ( similar to how you rationalize the denominator when it is surds)
• While you are doing the computation, always remind yourself that \[i^{2}=-1\]

3. Polynomial equations with real coefficients.

• For a polynomial equations with real coefficients, if z is a root, then z* is also a root.

If you want to know more about this theorem, you can refer to http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem

By now you should be able to do the whole tutorial 10a.