Complex+Numbers

__**Friday 22nd June**__

KEY LEARNING TODAY!!!

(i) j is the (a) square root of -1 ie j^2 = -1 (ii) a complex number can be written z = x + yj (iii) The REAL part of z is written Re(z) = x (iv) The IMAGINARY part of z is written Im(z) = y

__**ADDITION AND SUBTRACTION**__

Add/subtract the real parts, add/subtract the imaginary parts

eg (3 + 5j) + (4 - 2j) = 7 + 3j (8 + 3j) - (-1 + 11j) = 9 - 8j etc

Expand like you would a double bracket - remember that j^2 = -1
 * __MULTIPLICATION__**

eg (2 + 3j)(4 - j) = 8 - 2j + 12j - 3j^2 = 8 - 2j + 12j -3(-1) = 8 - 2j + 12j + 3 = 11 + 10j etc

__**SOLVING QUADRATIC EQUATIONS**__

Simply use the quadratic formula and use j when you take the square root of a negative number

eg the solutions to z^2 + 2z + 2 are (-2 ± sqrt(4 - 8)) / 2 = ............ = (-1 + j) and (-1 - j)

__**COMPLEX CONJUGATE**__

This is VERY important! z* is called the 'the complex conjugate of z' and z* = Re(z) - Im(z) (just change the sign of the imaginary part) So - if z = 4 + 3j then z* = 4 - 3j etc

NB (z + z*) = 2Re(z) and (z - z*) = 2Im(z)

By the end of the lesson we were happy working through exercise 2A (page 50)

Please make sure you are happy with EVERY question in that exercise!

We started today by noticing that if z = x + yj then... z multiplied by its conjugate = zz* =(x + yj) (x - yj)= x^2 + y^2 = = This has implications - we can 'realise' the denominator (we made up that expression!) for something like 1/(2 + 3j) by multiplying top and bottom by the complex conjugate of the bottom.
 * Monday 25th June**

ie 1/(2 + 3j) = 1/(2 + 3j) x (2 - 3j)/(2 - 3j) = (2 - 3j)/(4 + 9) = (2 - 3j) / 13

So...__**to DIVIDE by a complex number we can multiply top and bottom by the complex conjugate**__

eg (4 + 2j) ÷ (3 + 4j)

= (4 + 2j)/(3 + 4j) x (3 - 4j)/(3 - 4j)

= (4 + 2j)(3 - 4j)/(9 + 16)

= (12 - 16j + 6j + 8)/(25)

= (20 - 10j)/(25)

= (4 - 2j)/5

Also today - if two complex numbers z and w are equal then...Re(z) = Re(w) and Im(z) = Im(w)

This may seem obvious but it IS important...


 * HOMEWORK (due next Monday)**

The EVEN questions of exercise 2B (2, 4, 6, 8, 10)

__**Wednesday 27th June**__

Today's lesson:

Main ideas - we now know how to represent complex numbers on an ARGAND diagram. We understand the MODULUS and (PRINCIPLE) ARGUMENT of a complex number We understand the geometrical effect of complex conjugate, multiplying by j and -z

BE WARNED - you need to think about a complex number's QUADRANT when you work out its principle argument (you can use the POL function on your calculator)

__**Friday 29th June**__

Today's lesson:

There were principally two parts to today's lesson

(i) understanding that an equation like mod(z) = 3 or mod(z - (2 + 3j)) = 4 represents a circle. The first one has centre the origin and radius 3, the second one has centre (2 + 3j) and radius 4. Using inequality signs rather than equals means shading inside/outside the circle. PLEASE ensure that a strict inequality sign means a dotted boundary!!!

(ii) learning that if z is a root of ANY polynomial equation then z* is also a root. We will learn about properties of roots later on in the course that will make part (ii) of that exam question much more straightforward.

Homework is due in on Monday!

__**Monday 2nd July**__

Today's lesson:

PLUS: answers to the homework...:

Today we went through a bit of homework feedback - question 10 was a bit nasty but a nice challenge!

The rest of the lesson was all about modulus/argument form, plus a bit of exam practice. To recap: The modulus of z = x + yj = the distance of z from the origin = sqrt(x 2 + y 2 ) The (principle) argument is z is the angle (between pi and negative pi) you turn, anticlockwise from the positive x-axis to face z. Pol on your calculator will give the correct angle. Using tan -1 may give the WRONG angle - you NEED to consider the quadrant.

If that homework hasn't been submitted please do so asap!

__**Wednesday 4rd July**__ Today's lesson:

Today we looked at lines (or half-lines, if you prefer) and regions that satisfy equations like argz = (pi)/4 or arg (z - j) < -(pi)/4 etc

One thing to be careful about - an angle less than -pi/4 means 'greater rotation from the x-axis'...

__**Friday 6th July**__ Today's lesson:

Pretty much finished off complex numbers today.

(i) remember that if z is a root of a polynomial equation with real coefficients then z* is too (the complex conjugate) (ii) if z 1 is a root then (z - z 1 ) is a factor....we can use this to find ALL solutions to a polynomial equation if we are given a root! (up to degree 4 at any rate!)

__**Monday 9th July**__ I have looked through the past papers and here are the Complex Numbers questions that you should be OK with now..

June 2005: q2, q5, q9

Jan 2006: q2, q8 June 2006: q4, q8

Jan 2007: q2, q8 June 2007: q2, q4, q9

Jan 2008: q2, q3, q8 June 2008: q2, q9

Jan 2009: q1, q4, q9 June 2009: q2, q8

Jan 2010: q1, q8 June 2011: q3, q8

Jan 2011: q4 June 2011: q2, q8

Over the holidays please look at these. At the start of September I will take in..


 * June 2010: q3, q8**
 * Jan 2011: q4**
 * June 2011: q2 and q8**

FULL, WORKED SOLUTIONS PLEASE!!!


 * NOTE** we haven't covered the whole course yet and some questions (while doable now) will become simpler when we learn about sums/products of roots. So don't get too stressed out if you're not totally sure with all the answers yet!

Have a nice summer!