FP1: Chapter 1
Complex Numbers
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 2nd January 2014
i i captain!
Here’s something that someone has almost
certainly spoiled for you already…
𝑖 = −1
?
Complex versus imaginary numbers
Imaginary number: of form
𝑏𝑖,
?𝑏 ∈ ℝ
Complex number: of form
𝑎 + 𝑏𝑖,
? 𝑎, 𝑏 ∈ ℝ
Real
? part
Imaginary
? part
3 + 4𝑖
Putting complex numbers in the preferred form
We tend to ensure real and imaginary components are grouped together.
(a, b and c are real constants)
? + b)i
a + 3i – 4 + bi
= a – 4 + (3
2a – 3bi + 3 – 6ci = 2a+3 – 3(b
? + 2c)i
Convention 1
Just like we’d write 6 instead of 6, the i appears after any real constants, so
we might write 5ki or i.
An exception is when we involve a function.
e.g. i sin  and i√3
Why? This avoids ambiguity over whether ?
the function is being applied to the i.
Convention 2
In the same way that we often initially use x and y as real-valued variables, we
often use z first to represent complex values, then w.
Why complex numbers?
Complex numbers were originally introduced by the Italian mathematician Cardano in the 1500s to allow him to
represent the roots of polynomials which weren’t ‘real’. They can also be used to represent outputs of functions
for inputs not in the usual valid domain, e.g. Logs of negative numbers, or even the factorial of negative
numbers!
Some other major applications of Complex Numbers:
1
Fractals
A Mandelbrot Set is the most popular ‘fractal’. For
each possible complex number c, we see if zn+1 =
zn2 + c is not divergent (using z0 = 0), leading to the
diagram on the right. Coloured diagrams can be
obtained by seeing how quickly divergence occurs
for each complex c (if divergent).
2
This is called an Argand
Diagram. We’ll use it later.
Analytic Number Theory
Number Theory is the study of integers. Analytic Number Theory treats integers as
reals/complex numbers to use other (‘analytic’) methods to study them. For example,
the Riemann Zeta Function allows complex numbers as inputs, and is closely related to
the distribution of prime numbers.
3
Physics and Engineering
Used in Signal Analysis, Quantum Mechanics, Fluid Dynamics, Relativity, Control Theory...
Manipulation and application
Here’s a flavour of some of things you’ll be initially expected to do...
−36
? = 𝟔𝒊
= 𝟑𝟔 −𝟏
Solve 𝑥 2 + 9 = 0
𝒙𝟐 = −𝟗
𝒙 =? ± −𝟗
𝒙 = ±𝟑𝒊
Solve 𝑥 2 + 6𝑥 + 25 = 0
Using the quadratic formula:
𝒙 =?−𝟑 ± 𝟒𝒊
Simplify 5 + 2𝑖 + 8 + 9𝑖
= 13 + 11𝑖
?
Simplify (8 + i)(3 – 2i)
= 24 – 16i + 3i – 2i2
= 24 – 16i?+ 3i + 2
= 26 – 13i
Multiplying
Example: Let z1 = 1 + 3i and z2 = 5 – 2i, find z1z2.
(1 + 3i)(5 – 2i) = 5 – 2i + 5i – 6i2
The quick way to think about
this is that i2 reverses the sign.
=5 –?
2i + 5i + 6
= 11 + 3i
1 Evaluate the following:
a) √(-16)
= 4i ?
b) √(-25)
= 5i ?
c) √(-3)
= i√3 ?
d) √(-7)
= i√7 ?
e) √(-8)
=2i√2?
2 Calculate z2 for the following z.
a)
b)
c)
d)
e)
f)
z=1+i
z=1–i
z = 3 + 2i
z = 7 – 4i
z = -3 + 3i
z = a + bi
z2 = ?
2i
z2 = ?
-2i
z2 = 5?+ 12i
z2 = 33
– 56i
?
z2 = -18i
? 2
2
z = a2 –?b + 2abi
3
Calculate z1z2.
a) z1 = 1 + i
b) z1 = 2 + i
c) z1 = 3 + 2i
d) z1 = -3 + i
e) z1 = 7 – 3i
f) z1 = -1 – i
z2 = 1 – i
z2 = 2 – 2i
z2 = 4 – 3i
z2 = 5 + 7i
z2 = 3 – 7i
z2 = 9 + 8i
z1z2 =?2
z1z2 = 6?– 2i
z1z2 = 18
?–i
z1z2 = -22
? – 16i
z1z2 = -58i
?
z1z2 = -1
? – 17i
4
By using a Binomial Expansion,
determine (1 + i)5.
= 1 + 5i + 10i2 +10i3 + 5i4 + i5
= 1 + 5i – 10 –?10i + 5+ i
= -4 – 4i
5
What is the value of:
a) i100 = (i4)25 =?125 = 1
b) i2003 = i2000 x i3 ?
= 1 x i3 = -i
Complex Conjugates
Remember in C1 how we rationalised the denominator?
3
4− 2
Suppose x1 = a + √b and x2 = a - √b
Then:
x1 x2 = a 2 – b
x1 + x2 = 2a
Both these results are rational.
?
The same trick works with complex numbers.
! If z = x + yi, then we define z* = x – yi.
z* is known as the complex conjugate of z.
z z* = x2 + ?
y2
z + z* = 2x ?
which are both real.?
The fact the first is real
will help us with
dividing complex
numbers.
We’ll see the
significance of the
second later.
Quickfire Questions
Given z, determine z z*.
1
2
3
4
5
z = 3 + 2i
z = 3 – 2i
z = 5 + 4i
z=1+i
z = 4 – 2i
z z* = 13?
z z* = 13?
z z* = 41?
z z* = 2?
z z* = 20?
Dividing
__26__
(2 – 3i)
(2 + 3i )(2 – 3i)
=
52 – 78i
13
=
4 – 6i
Click to Brosolve
Exercises
Put all the following in the form a + bi.
1
2
3
4
_1_
1+i
_10_
3+i
1-i
1+i
= 1?- i
2
Note that mark
schemes permit the
single fraction.
5
= 3 ?- i
= -i ?
14 – 5i = 4 ?+ i
3 – 2i
9
10 + 3i = 16 –?17i
1 + 2i
5
6
_i_
1–i
= -1 + i?
2
7
8 – 4i
1 – 3i
= 2 + 2i
?
8
a+i
a–i
2 – 1 + 2ai
a
=
?
2
a +1
Edexcel June 2013 (Retracted)
Given that (2 + i)(z + 3i) = 10 – 5i, find z,
giving your answer in the form a + bi.
z = 3 – 7i (either by dividing 10 – 5i by 2 + i and
subtracting 3i, or replacing z with a + bi before
expanding and comparing real and imaginary parts).
?
Argand Diagrams
Argand Diagrams are a well of geometrically representing complex numbers.
Im[z]
Click to move.
3
1+i
2
2 – 3i
1
-3
-2
-1
1
2
3
-1
-2
-3
z = x + yi
The x-axis is the real component.
The y-axis is the imaginary component.
Re[z]
-1 + 2i
-3 – i
Argument and Modulus
In FP2, you’ll encounter something called ‘polar coordinates’. This is an alternative
way of representing coordinate, which instead of using the (x,y) position (known as a
Cartesian coordinate), uses the distance from the origin and the angle.
(Don’t write anything down yet!)
Im[z]
z = 2 + 3i
3
Distance from origin:
2
1
-3
-2
-1
-1
-2
-3
?

1
√(22 + 32) = √13
This is known as the
modulus of z, and we write
|2 + 3i| = √13
2
3
Re[z]
Angle: (anticlockwise from the real axis)
Using trigonometry, we
have (in radians):
tan-1(3/2) = 0.983
This is known
? as the
argument of z, and we
write:
arg(2 + 3i) = 0.983
Argument and Modulus
Im[z]
!
If 𝑧 = 𝑥 + 𝑦𝑖
arg z
Re[z]
Modulus of z:
𝒛 = 𝒙 𝟐 + 𝒚𝟐
Argument of z:
Usual range - < arg z ≤ . Known as the
principal argument if in this range.
−𝟏
𝒂𝒓𝒈 𝒛 = 𝒕𝒂𝒏
𝝅
𝒚
𝒙
𝝅
when − < 𝒂𝒓𝒈 𝒛 <
𝟐
𝟐
Use trig common sense for other
quadrants.
Check your Understanding
z
|z|
1
1
-i
1
-1 + i
√2
-5 – 2i
√29
arg(z)
?
?
?
?
0
-/2
3/4
-2.761
?
?
?
?
Exercises
Give exact answers where possible, otherwise to 3dp.
1
z
|z|
arg(z)
-1
1
i
1
1+i
√2
1 + 2i
√5
1 – 2i
√5
-1 + 2i
√5
-1 – 2i
√5
3 + 4i
5
-5 + 12i
13
1 – i√3
√7
?
?/2
?p/4
?1.107
?-1.107
?2.034
?-2.034
?0.927
?1.966
?-/3
2
z
|z|
arg(z)
-√3 - i
√7
3+i
√10
2 – 5i
√29
-4 + 3i
5
-1 – 4i
√17
?-5/6
?0.322
?-1.190
?2.498
?-1.816
3
Given that arg(3 + a + 4i) = p/3 and that
a is real, determine a.
4/(3 + a) = √3. Thus
? a = (4/√3) – 3
4
Give that arg(5 + i + ai) = p/4 and that a
is real, determine a.
(a + 1)/5 = 1. So a?= 4
5
Given that arg(santa + 3 + 2i) = /2 and
that santa is real, determine santa.
santa = -3
?
Argument-modulus form
What’s the point of |z| and arg(z)?
We’ll see in FP2 that we can express a complex number as 𝑧 𝑒 𝑖 arg
see the infamous 𝑒 𝜋𝑖 = −1. But for now, let’s get halfway there...
Im[z]
𝑧
., where we’ll also
Suppose that r = |z| and  = arg(z).
How could we express z in terms of
r and o?
arg z
Re[z]
!
z = x + yi
= r cos  + r i sin 
= r(cos  + i sin )
This is known as the modulusargument form (or the polar form)
of z.
Argument-modulus form
Question 1 from earlier...
1
z
|z|
arg(z)
Mod-arg form
-1
1

z = cos  + i sin ?
i
1
/2
z = cos(/2) + i sin(/2)
?
1+i
√2
p/4
z = √2(cos(/4) ?
+ i sin(/4))
1 + 2i
√5
1.107
z = √5(cos 1.107?+ i sin 1.107)
1 – 2i
√5
-1.107
z = √5(cos(-1.107)
? + i sin(-1.107))
-1 + 2i
√5
2.034
z = √5(cos 2.034?+ i sin 2.034)
-1 – 2i
√5
-2.034
z = √5(cos(-2.034)
? + i sin(-2.034))
3 + 4i
5
0.927
z = 5(cos 0.927 +?i sin 0.927)
-5 + 12i
13
1.966
z = 13(cos 1.966+
? i sin 1.966)
1 – i√3
√7
-/3
z = √7(cos(-/3)?+ i sin(-/3))
|z1z2| and arg(z1z2)
When we multiply two complex numbers, we multiply their moduli, and we add their
arguments.
So can we prove this, i.e. that:
𝒛𝟏 𝒛𝟐 = 𝒛𝟏 𝒛𝟐
and
𝐚𝐫𝐠 𝒛𝟏 𝒛𝟐 = 𝐚𝐫𝐠 𝒛𝟏 + 𝐚𝐫𝐠 𝒛𝟐
Let 𝒛𝟏 = 𝒓𝟏 𝐜𝐨𝐬 𝜽𝟏 + 𝒊 𝐬𝐢𝐧 𝜽𝟐 and 𝒛𝟏 = 𝒓𝟏 𝒄𝒐𝒔 𝜽𝟏 + 𝒊 𝒔𝒊𝒏 𝜽𝟐 .
Then 𝒛𝟏 𝒛𝟐 = 𝒓𝟏 𝒓𝟐 𝐜𝐨𝐬 𝜽𝟏 𝐜𝐨𝐬 𝜽𝟐 − 𝐬𝐢𝐧 𝜽𝟏 𝐬𝐢𝐧 𝜽𝟐 + 𝒊 𝐬𝐢𝐧 𝜽𝟏 𝒄𝒐𝒔 𝜽𝟐 + 𝒔𝒊𝒏 𝜽𝟏 𝐜𝐨𝐬 𝜽𝟐
= 𝒓𝟏 𝒓𝟐 𝐜𝐨𝐬 𝜽𝟏 + 𝜽𝟐 + 𝒊 𝐬𝐢𝐧 𝜽𝟏?+ 𝜽𝟐
So 𝒛𝟏 𝒛𝟐 = 𝒓𝟏 𝒓𝟐 = 𝒛𝟏 𝒛𝟐
and 𝒂𝒓𝒈 𝒛𝟏 𝒛𝟐 = 𝜽𝟏 + 𝜽𝟐 = 𝐚𝐫𝐠 𝒛𝟏 + 𝐚𝐫𝐠 𝒛𝟐
The following C3 identities may be useful:
cos 𝑎 + 𝑏 = cos 𝑎 cos 𝑏 − sin 𝑎 sin 𝑏
sin 𝑎 + 𝑏 = sin 𝑎 cos 𝑏 + cos 𝑎 sin 𝑏
Comparing coefficients
If two complex numbers are equal, then clearly both their real and imaginary
parts are equal. i.e. if 𝑎1 + 𝑏1 𝑖 = 𝑎2 + 𝑏2 𝑖 then 𝑎1 = 𝑎2 and 𝑏1 = 𝑏2 .
Q Given that 3 + 5𝑖 = (𝑎 + 𝑖𝑏)(1 + 𝑖) where 𝑎 and 𝑏 are real,
find the value of 𝑎 and 𝑏.
Expanding: 𝒂 − 𝒃 + 𝒊 𝒂 + 𝒃 = 𝟑 + 𝟓𝒊
So 𝒂 − 𝒃 = 𝟑 and 𝒂 + 𝒃 = 𝟓.
?
Solving: 𝒂 = 𝟒, 𝒃 = 𝟏. We could have also found
Q Calculate 𝑖.
𝒊 = 𝒂 + 𝒊𝒃
𝒊 = 𝒂 + 𝒊𝒃 𝟐 = 𝒂𝟐 + 𝟐𝒂𝒃𝒊 − 𝒃𝟐
So 𝟎 + 𝟏𝒊 = 𝒂𝟐 − 𝒃𝟐 + 𝟐𝒂𝒃 𝒊
𝒂𝟐 − 𝒃𝟐 = 𝟎 and 𝟐𝒂𝒃 = 𝟏
𝟏
𝒂=𝒃=
𝟐
𝟏
𝟏
So 𝒊 = + 𝒊
?
𝟐
𝟐
𝟑+𝟓𝒊
𝟏+𝒊
Exercise 1G
1
𝑎 + 2𝑏 + 2𝑎𝑖 = 4 + 6𝑖 where 𝑎 and 𝑏 are real.
Find the value of 𝑎 and of 𝑏.
𝟏
𝒂 = 𝟑, 𝒃 =
𝟐
?
4
7
𝑎+𝑖
𝒂=𝟑
3
= 18 + 26𝑖 where 𝑎 is real. Find the value of 𝑎.
?
1
Find real 𝑥 and 𝑦 such that
= 3 − 2𝑖
𝑥+𝑖𝑦
𝟑
𝟏
𝒙 = ,𝒚 = −
𝟐
𝟐
?
9
Find the square roots of 7 + 24𝑖
± 𝟒 + 𝟑𝒊
?
10 Find the square roots of 11 + 60𝑖
±(𝟔 + 𝟓𝒊)
?
11 Find the square roots of 5 − 12𝑖
± 𝟑 − 𝟐𝒊
?
Roots of polynomials
Any cubic (with coefficient of x3 of 1) can be expressed as:
y = (x – α)(x – )(x – )
where α,  and  are the roots.
However, these 3 roots may not necessarily all be real...
α

All 3 roots are real.

α
α
1 real root, 2 ?
complex roots.
3 real roots (one
? repeated)
Are there any other possibilities?
No: since cubics have a range of – to +, it must cross the x axis.
And it can’t cross an even number of times, otherwise the cubic
would start and end in the same vertical direction.
?
=
Roots of polynomials
Classic Question:
x = 2 is one of the roots of the polynomial x3 – x – 6
Find the other two roots.
Use polynomial division (as per C2) to divide x3 – x – 6 by (x – 2).
This gives x2 + 2x + 3.
?
Using the quadratic formula, we obtain the roots:
x = -1  i√2
Your Go:
x = -1/2 is one of the roots of the polynomial 2x3 – 5x2 + 5x + 4.
Find the other two roots.
We could divide by (x + ½), but it would be cleaner to divide by (2x + 1).
This gives x^2 – 3x + 4. Using the quadratic formula, we obtain the roots:
?
x = - ½ (3  i√7)
Relationship between complex roots
For the first cubic, we found the two complex roots were:
x = -1  i√2
Q
What is the relationship between these two roots?
They’re complex conjugates.
?
Q
Can we prove this will always be the case for cubics with real coefficients?
Suppose a and b are the two complex roots. Then (x – α)(x – β) in the
expansion must give real coefficients.
(x – α)(x – β) = x2 – (a + β)x + αβ
Therefore α + β and αβ must both be real. We saw earlier that this is
satisfied when α and β are complex conjugates.
?
(Although technically our proof isn’t complete, because we’ve shown that the roots being
complex conjugates is a sufficient condition for the coefficients of the cubic being real, not a
necessary one. However, we know the imaginary components of α and β must be the same but
negative of each other, so that in α + β they cancel out, and we can proceed from there to show
the real components of α and β must be the same).
In summary, complex roots of polynomials always come in
complex conjugate pairs.
Finding other roots
Another classic type of exam question:
-1 + 2i is one of the roots of the cubic x3 – x2 – x – 15.
Find the other two roots.
Other complex root is the complex conjugate: -1 – 2i.
Now expand (x – (-1 + 2i))(x – (-1 – 2i))
= x2 – (-1 – 2i)x – (-1 + 2i)x + (-1 + 2i)(-1 – 2i)
= x2 + 2x + 5
?
We can now either use polynomial division or “Factor
Theorem with trial and error” to establish the real root as 3.
Your turn:
2 – i is one of the roots of the cubic x3 – 11x + 20.
Find the other two roots.
2 + i is other complex root.
(x – (2 + i))(x – (2 – i)) = x2 – 4x
? + 5.
Dividing we get (x + 4), so real root is -4.
Quartics
Any quartic (with coefficient of x4 of 1) can be expressed as:
y = (x – α)(x – )(x – )(x – δ)
where α, ,  and δ are the roots.
α


δ
Possibility 1:
4 real roots (some
?
potentially repeated)

α
Possibility 2:
2 real roots, pair of
? roots.
complex conjugate
Possibility 3:
No real roots. Two pairs
of complex ?
conjugate
roots.
Any other possibilities?
No. On y-axis, both ends of line must either be both positive infinity or both negative. xaxis must therefore be crossed an even number of times, so an even number of real roots.
?
Quartics
Edexcel Jan 2013 (Retracted)
a) 4x2 + 9 = 0
x2 = -9/4,
x2
x = i√(3/2)
b)
Im[z]
Since real roots appear
on the x-axis and complex
roots come in conjugate
pairs, we know we’ll have
symmetry in the line y = 0
?
– 2x + 5 = 0
Using the quadratic formula, x = 1  2i
?
Re[z]
Exercise 1H
1
Given that 1 + 2i is one of the roots of a quadratic equation with real coefficients, find the
equation.
𝒙𝟐 − 𝟐𝒙 + 𝟓 = 𝟎
?
3
Given that 𝑎 + 4𝑖, where 𝑎 is real, is one of the roots of a quadratic equation with real
coefficients, find the equation.
𝒙𝟐 − 𝟐𝒂𝒙 + 𝒂𝟐 + 𝟏𝟔 = 𝟎
?
5
Show that 𝑥 = 3 is a root of the equation 2𝑥 3 − 4𝑥 2 − 5𝑥 − 3 = 0.
Hence solve the equation completely.
𝟏
𝟏
𝟏
𝟏
Roots are 3, − + 𝒊, − − 𝒊
?𝟐
7
𝟐
𝟐
𝟐
Given that −4 + 𝑖 is one of the roots of the equation 𝑥 3 + 4𝑥 2 − 15𝑥 − 68 = 0, solve
the equation completely.
Roots are 4, -4 + i and -4 - i
?
9
Given that 2 + 3𝑖 is one of the roots of the equation 𝑥 4 + 2𝑥 3 − 𝑥 2 + 38𝑥 + 130 = 0,
solve the equation completely.
Roots are 2 + 3i, 2 – 3i, -3 + i and -3 – i
?
4
10 Find the four roots of the equation 𝑥 − 16 = 0. Show these roots on an Argand Diagram.
?
Roots are 2, -2, 2i, -2i