1A_Maths_C_V

3. Complex Variables

Complex numbers were introduced as an extension of the real number system so that every
polynomial equation would have solutions.

e.g. x2 + 5 = 0 has no real solution.

Originally something of a curiosity, over the years it has been found that a good many engineering

problems (the first one you will meet is a.c. circuit theory) are easier to solve when considered as a

problem for a complex quantity rather than for a real one. x = ±√5 i

3.1 Basic Properties

Introduce i = –1 , and represent a general complex number as z = x + i y. The traditional way
of representing z graphically is the Argand diagram/plane.

Im(z) z = x + i y is the cartesian form, we also have the
polar form,
z
z = r cosθ + i r sinθ
r
r is the modulus of z, denoted | z |
y θ is the argument or phase of z,
denoted arg(z) or ph(z).
θ

x Re(z)

We can move easily between these the cartesian (x,y) and the polar (r,θ) form of z:

x = r cos θ r = z = x2 + y2

y = r sin θ θ = tan−1 y ( + π if x is negative)
x

For all reasonable arithmetic operations, complex numbers can be manipulated in exactly the same

way as reals,

z1 + z2

Addition z1 + z2 = ( x1 + i y1 ) + ( x2 + i y2 ) z2
= ( x1 + x2 ) + i ( y1 + y2 )
z2

(addition of complex numbers is as for vectors) z1
Subtraction z1 − z2 = ( x1 + i y1 ) − ( x2 + i y2 )
N.B. Error on handout
= ( x1 − x2 ) + i ( y1 − y2 )

Standard Pace Maths Course Complex Variables 1

Multiplication z1 z2 = ( x1 + i y1 ) ( x2 + i y2 )

= x1x2 − y1 y2 + i ( x1 y2 + x2 y1 )

E.g. (5 + 2 i )(1+ i ) = 5 − 2 + i ( 5 + 2 ) = 3 + 7i

Division z1 = x1 + i y1
z2 x2 + i y2

We usually want the answer in the form a + ib. To find this, we use the trick of multiplying top and
bottom by something to make the denominator real. This is x2 − i y2 to give

z1 = x1 + i y1 x2 − i y2
z2 x2 + i y2 x2 − i y2

= ( x1x2 + y1 y2 ) + i ( y1x2 − x1y2 )

x22 + y22

E.g. Express 3 − 2i in the form a + ib
1+ i

3 − 2i = 3 − 2i 1 − i = 1− 5i = 1 − 5 i
1+ i 1+i 1−i 12 +12 2 2

3.2 Complex Conjugation Im(z)

Definition z
The complex conjugate of z = x + i y is x − i y
y
The complex conjugate is denoted z
(or sometimes z* ) −y Re(z)
z
e.g. If z = 3 + 4i, then z = 3 − 4i

Note that on the Argand diagram complex conjugation is
equivalent to a reflection of a number in the real axis.

Useful Properties of complex conjugation z2
(i) z z is a real number. It is equal to the square of the modulus i.e. z 2

If z = x + iy, then z z = ( x + iy)( x − iy) = x2 + y2 =

(ii) Re(z) = z + z and Im(z) = z − z (Put z = x + iy and check these).
22

(iii) If z = z , then z is real.

Standard Pace Maths Course Complex Variables 2

(iv) z = z
This is obvious geometrically.

(v) Conjugation “passes through” basic arithmetic e.g

z1 + z2 = z1 + z2 Exercise: write z1 = x1 + iy1 , z2 = x2 + iy2 and check these.
z1z2 = z1 z2 The essence of this property is that to conjugate an expression,
z1 = z1 we simply need to change all the i’s to −i’s.
z2 z2

3.3 The exponential form z = r eiθ

3.3.1 Defining functions of complex numbers

We want to define functions such as ez and sin z, where z is a complex number. A good way

to do this is using power series. For real x we have
exp( x ) = ex = 1 + x + x2 + x3 + x4 + ...
2! 3! 4!

For complex arguments, i.e. with z = x + iy

Define exp( z ) = ez = 1+ z + z2 + z3 + z4 + ... valid for all z
2! 3! 4!

Note that the answer we obtain for ez will also in general be a complex number.

The familiar properties of the exponential function are exactly the same for complex arguments as
for real ones, e.g.

ew ez = ew+ z ( ez )α = eα z

Similarly we can define

sin z z3 z5 valid for all z
= z – + – ...
3! 5!

cos z z2 z4
= 1 – + – ...

2! 4!

z2 z3 1
z– +
ln (1 + z) = 23 – ... valid for | z | < 1

Same radius of convergence as real case

ln (1 + x) = x – x2 + x3 – ... valid for | x | < 1

23

Standard Pace Maths Course Complex Variables 3

3.3.2 Euler’s Formula: eiθ = cos θ + i sin θ

If you only remember one thing about complex numbers, this should be it. Using our definition

above for ez , we have

eiθ = 1 + iθ + (iθ )2 + (iθ )3 + (iθ )4 + ... i2 = −1
2! 3! 4! i3 = −i
i4 = +1
= 1+ iθ − θ2 − i θ3 + θ4 + i θ5 + ... i5 = +i
2! 3! 4! 5!

= 1− θ2 + θ4 + ... + iθ − iθ33! + i θ5 + ...
2! 4! 5!

i.e.

eiθ = cosθ + isinθ = Euler's formula

What does this mean? Since cos2θ + sin2θ = 1, then the eiπ/2 i
modulus of this complex number is 1, i.e. eiθ lies on the
unit circle. As θ varies, eiθ traces out the whole of the unit eiθ
circle.
eiπ θ1
Examples −1 ei2π

eiπ / 2 = cosπ 2 + i sinπ 2 = i ei3π/2 −i
eiπ = cosπ + i sin π = −1

ei3π / 2 = cos 3π 2 + i sin 3π 2 = − i = e−iπ / 2

e2iπ = cos 2π + i sin 2π = 1

e2iπ n = 1 where n is an integer – this one is useful for simplifying expressions

3.3.3 Multiplication in exponential form

Recall that we could express z in polar form as r cos θ + i r sin θ . We now see from Euler's

formula that we can write z as z = reiθ

z = r eiθ r

θ

This is called the exponential form of a complex number. Since ew ez = ew+ z , multiplying
complex numbers is easier to understand:

Standard Pace Maths Course Complex Variables 4

If z1 = r1 eiθ1 and z2 = r2eiθ2 then

( )z1 z2 = r1 eiθ1 r2 eiθ2 = r1 r2 ei θ1+θ2

so when we multiply two complex numbers, we

• multiply the moduli together i.e. z1 z2 = z1 z2
• and add the arguments i.e. arg ( z1 z2 ) = arg ( z1) + arg ( z2 )

similarly when we divide complex numbers, we

• divide the moduli i.e. z1 = z1

z2 z2

• and subtract the arguments i.e. arg ( z1 / z2 ) = arg ( z1 ) − arg ( z2 )

This means that multiplying a complex number w by another one z = r eiθ , on the Argand

diagram the effect is equivalent to stretching w by a factor of r and rotating it by θ.

Examples

(i) What is 1+ i ?
2 + 3i

Using the last statement 1+ i = 1+ i = 2
2 + 3i 2 + 3i
.
13

Checking this directly 1+ i = 1+ i 2 − 3i = 5 − i and 1+ i = 26 = 2
2 + 3i 2 + 3i 2 − 3i 13 2 + 3i 13 13

(ii) 2 i = 2eiπ 2 in exponential form, so has modulus z
2 and argument π / 2.
So multiplying by 2 i doubles the length and rotates 2i z
through π / 2 radians 2| z |

2i z = 2eiπ / 2 r eiθ = 2r ei(θ +π / 2)

Standard Pace Maths Course Complex Variables 5

(iii) Multiply 1 + i and 1 + i in exponential form and convert back Im z

1 + i = 2 eiπ 4 √2 1
( 1 + i ) ( 1 + i ) = 2 eiπ 2 = 2i π/4

N.B. Note that this example is artificial – if numbers are given in 1 Re z
Cartesian form x + iy, we would not normally convert them to

exponential form to do multiplication. Before diving in, it is always

worth considering which form (Cartesian, polar or exponential) is likely to lead to the easiest

expressions. Remember also that your calculator has complex mode and will move back and forth

between the various forms

(iv) Show that

cosθ = eiθ + e−iθ and that sinθ = eiθ − e−iθ
2 2i

eiθ = cosθ + isinθ (= cos(−θ ) + isin(−θ ))
e−iθ = cosθ − isinθ

Adding and subtracting gives eiθ + e−iθ = 2 cosθ and eiθ − e−iθ = 2i sinθ

(v) Find the complex number which when multiplying other complex numbers has the effect of
rotating clockwise by 60˚ and shrinking length by a factor of 2.

z = 1 e−iπ 3 = 1 cos π − i sin π
2 2 3 3
Careful – Euler's Formula
= 1 1 −i 3 = 1 −i 3 valid in radians

22 2 44

(vi) What is e2+i ?

This looks tricky until you realize that we can separate out the real and imaginary parts of the
exponent

( )e2+i = e2ei = 7.39 cos1+isin1 = 3.99 + 6.22i

(vi) What is 1 in polar form ?
z

With z = r eiθ , 1 = 1 = 1 e−iθ
z r eiθ r

Standard Pace Maths Course Complex Variables 6

3.3.4 de Moivre's Theorem

This is a useful trigonometric identity follows directly from the properties of the exponential

function. Since einθ = ( eiθ )n then

cos nθ + i sin nθ = (cos θ + i sin θ )n

Example
Use de Moivre’s theorem to find express cos 2θ and sin 2θ in terms of cosθ and sinθ .

( )cos 2θ + i sin 2θ = cosθ + isinθ 2

= cos2θ + i 2sinθ cosθ − sin2θ

So comparing real and imaginary parts, we have sin 2θ = 2sinθ cosθ
cos 2θ = cos2 θ − sin2 θ

3.4 Principal value of θ

Im(z) For any integer, n,

z e 2  in = cos 2π n + i sin 2π n = 1

r Thus for any z = r eiθ we can add 2n¡ to θ and in

θ no way affect the value of z.

θ + 2π Re(z) ( )z = r eiθ = reiθ ei2nπ = rei θ +2nπ

Usually it does not matter in practice that many

values of θ can represent a given z. There are, however, a few instances when these things are
crucial (see for example section 3.6 on calculating nth roots). Sometimes, to avoid ambiguity in θ,

we use the principal value of θ, with θ restricted to the range

−π < θ ≤ π

It is then usual to interpret, unless told otherwise, that a potentially multiple-valued function is based

on the principal value of θ.

Summary Don't start Examples 4/3 yet

• i = −1 behaves just like any number in all algebraic manipulation and i2 = −1

• Need to be familiar with the properties of ( ) , | ( ) | and arg( ), such as

z1z2 = z1 z2 z1z2 = z1 z2 arg ( z1 z2 ) = arg ( z1) + arg ( z2 )

• eiθ = cosθ + i sinθ • cos nθ + i sin nθ = (cos θ + i sin θ )n

Standard Pace Maths Course Complex Variables 7

3.5 Plotting Graphs

When designing circuits and/or filters, or in measuring the stability of a system, equations involving
relationships between the moduli of various complex numbers often crop up and these are easy to
solve use geometric reasoning, provided you can sketch/plot them.

Example z
Find the locus of points on the Argand diagram which consists |z| |z−3|

of points z which satisfy

|z| = 2|z−3|

Before diving in, it is useful to think a little about what this means 23 6

| z | is the distance of the point z from the origin and

| z − 3 | is the distance of z from 3

There is a point at 2 and a point at 6 , we should have symmetry in y and the locus is probably a

bounded curve i.e. it can not get very far away from the origin. If a point was far away, then the

distances from the origin and from 3 would be about the same.

Substituting the formula for the modulus 4+2eiθ
z = x2 + y2 2θ
4
z − 3 = x − 3+ iy = ( x − 3)2 + y2

Squaring both sides, therefore, gives

x2 + y2 = 4 x2 − 6x + 9 + y2

3x2 − 24x + 36 + 3y2 = 0
x2 −8x +12 + y2 = 0

( x − 4)2 + y2 = 4 i.e. a circle centre (4,0) and radius 2

There are a whole variety of ways circles can be represented in complex form.

Another two would be (b) z = 4 + 2eiθ −π < θ ≤ π

(a) z − 4 = 2

Exercise
Show that this locus can also be written

z z − 4 (z + z) + 12 = 0

Standard Pace Maths Course Complex Variables 8

3.6 nth roots
We can calculate the cube root of a complex number z = r eiθ , as follows

z1/3 = reiθ 1/ 3 = r1/ 3eiθ / 3

However, remember that if z = r eiθ , then z = r eiθ +i2π n as well, since ei2π n = 1. If we had

chosen to represent z as z = r ei(θ +2π ) then we could find a different cube root

z1/3 = reiθ +i2π 1/ 3 = r1/ 3eiθ / 3+i2π / 3

If instead we had chosen z = r ei(θ+4π) then we would get

z1/3 = reiθ +i4π 1/ 3 = r1/ 3eiθ / 3+i4π / 3

Every time we add a multiple of 2¡ to the argument of z, the argument of z1/3 changes by 2π/3 .

If we had tried the next representation z = r ei(θ+6π) , then

z1/3 = reiθ +i6π 1/ 3 = r1/ 3eiθ / 3+i6π / 3 = r1/ 3eiθ / 3+i2π

which is the same as our first cube root. The most we can ever find is 3 distinct cube roots.

E.g. Solve z 3 = −1 . (1)

Taking only a single value for the argument of –1 leads to a single root for z,

z 3 = –1 = eiπ z = eiπ / 3

( )eiπ/3 or z = ei π / 3+2π / 3 = eiπ

2π/3 1 π/3 ( )or z = ei π / 3+4π / 3 = ei5π / 3 = e−iπ / 3

Converting to cartesian form

eiπ z = cos π + i sin π = 1 + i 3

3 322

e−iπ/3 or z = cosπ + i sin π = −1

or z = cos − π + i sin − π = 1 − i 3
3 3 22

This approach extends to nth roots. One nth root of z = r eiθ is r1/ neiθ / n and the others are

r1/ nei(θ +2π ) / n , r1/ nei(θ +4π ) / n , ... , r1/ nei(θ +2(n−1)π ) / n . These are at the vertices of a regular

n-gon.

Standard Pace Maths Course Complex Variables 9

3.7 Polynomials with real coefficients

Roots of polynomial equations with real coefficients are either real or appear as complex

conjugate pairs. (Remember the definition of complex conjugation in section 3.2 and its properties,
e.g. if a is real then a = a ).

Theorem

If p(z) = an z n + an–1 z n–1 + .... + a1 z + a0 ( a's real )

Then the roots of p(z) = 0 are either (i) real or (ii) complex conjugate pairs.

Proof: The fundamental step in this proof is to show that, if z1 is a root, then so is z1 .

Let z1 be a root ,

an z1n + an−1 z1n−1 + ... + a1 z1 + a0 = 0

( )an z1n + an−1 z1n−1 + ... + a1 z1 + a0 = 0 taking of both sides

an z1n + an−1 z1n−1 + ... + a1 z1 + a0 = 0 z2 + z3 = z2 + z3

an z1n + an−1 z1n−1 + ... + a1 z1 + a0 = 0 z2z3 = z2 z3

n −1 a’s real and zn = z n

z1
( )( ) ( )an n +
an −1 + ... + a1 z1 + a0 = 0
z1

i.e. z1 is a root.

So either z1 = z1 (i.e. the root is real), or z1 and z1 are both (distinct) roots.

Example

Given that z = 1 + i is a root of the equation

z 4 − z 3 − 6 z 2 + 14 z − 12 = 0

Find the others.

Real coefficients z = 1 − i is also a root

( )Hence z − (1+ i) and z − 1− i are factors of the LHS N.B. Error on handout

( ) ( )z − 1+ i z − 1−i = z2 − 2z + 2 is a factor of the LHS

Finding the other quadratic factor (could do by inspection or by long division)

0 = z4 − z3 − 6z2 +14z −12 = z2 − 2z + 2 z2 + z − 6

= z2 − 2z + 2 (z +3)(z − 2)

z = 1 ± i , −3 and 2. Complex Variables 10
Standard Pace Maths Course

3.8 Complex Trigonometric and hyperbolic functions
Recall cosh x = e x + e−x for real x. (Unsurprisingly) we extend this definition to complex

2

numbers i.e. for complex z we define

cosh z = ez + e−z (1)
2

We saw in example (iv) on page 6 that

cos z = ( eiz + e−iz 2 (2)
2

Setting z = iw in (1) we see that cosh iw = cos w

If instead we try z = iw in (2) we obtain cos iw = cosh w

What about sin x and sinh x? Repeating a similar line of analysis:

sinh z = ( ez − e−z (3)
2

sin z = ( eiz − e−iz i (4)
2i

Setting z = iw in (3) and (4) we see that sinh iw = i sin w sin iw = i sinh w

These are very handy results since, by combining them with the addition formulae for trigonometric
functions (e.g. sin(A + B) = sin A cos B + cos A sin B – see page 2 of the Data book for more of
these), we have a handy way to calculate trigonometric functions of complex arguments

Examples
(i) What are the real and imaginary parts of sin(x + iy) ?

sin (x + iy) sin ( x + iy) = sin xcosiy + cos xsin iy
= sin xcosh y + icos xsinh y

Standard Pace Maths Course Complex Variables 11

(ii)* (=Tripos standard). What is sin-1 2 ?

This must be complex (since | sin x | ≤ 1 for real x), so we start by letting it = x + iy (x and y real)

2 = sin ( x + iy) = sin xcosh y + i cos xsinh y using example (i)

Equating real and imaginary parts gives

sin x cosh y = 2 and cos x sinh y = 0

If, in the second equation, we try sinh y = 0, then cosh y = 1 and we can not solve the

first equation. We can only solve the second equation, then, by taking

cos x = 0 x = ± π + 2kπ sin x = ±1 cosh y = ± 2
2

and we see that the − sign doesn't work.

The solution is, therefore, z = π + 2kπ ± i cosh−12 k an integer
2

Note that the ± sign has crept back in because cosh is an even function.

3.9 ln z cosh y = 2 y = ± cosh−12

The natural log, ln z , can be manipulated as follows

( )ln z = ln( reiθ ) = ln r + ln eiθ = ln r + iθ

Clearly the value chosen for θ influences the value of ln z directly. Since we can add 2nπ to the

value of θ without affecting z, we can add 2nπi to the value of ln z . ln z is an important function
since it is how we give meaning to the expression zw when w is complex. We define

zw = eln zw = ewln z

Example

Find ln ( 2 – i ).

2−i = reiθ with r = √5 and θ= tan−1 − 1 = −0.464
2

( )So that ln ( 2 – i ) = ln 5 − i0.464 + 2nπ i

Summary

• Roots of polynomials with real coefficients are real or complex conjugate pairs.
• Can add multiples of 2π to arg(z) – this matters for finding: roots, ln z , etc.
• Trigonometric and hyperbolic functions are related by sin iz = i sinh z, etc (in Databook)

Can now do Examples 4/3 Q1-6

Standard Pace Maths Course Complex Variables 12

3.10 Make it complex, make it easier.

We have seen that trigonometric functions are actually complex exponentials

eiθ = cosθ + i sinθ cosθ = eiθ + e−iθ and sinθ = eiθ − e−iθ
2 2i

Whenever a problem involving trigonometric functions becomes tedious, we always have the option
of recasting it as one involving complex exponentials, which often produces significant
simplifications.

Example (i) π e−2x cos x dx
Find I =
0

Using complex exponentials

I= π e−2x eix + e−ix dx
02

= 1 π e−2x +ix dx + 1 π e−2x−ix dx = 1 − e−(2−i)x − e−(2+i)x π
20 20 0
2 2−i 2+i

= 1 − e−(2−i)π −1 − e−(2+i)π −1 = 1 e−2π + 1 + e−2π + 1
2 2−i 2+i 2 2−i 2+i

( )=
e−2π + 1 4 = 2 e−2π + 1 eiπ = e−iπ = −1
2 22 +1 5

It is worth stepping back from this for a moment and noticing that at each stage of the manipulation,
the terms arising from the two integrals are complex conjugates of each other. They started off
being the same “except for the sign of i” and at each stage this remained true. This suggests that we
could have saved half the effort by only doing one of them.

We can accomplish this by noting that cos θ = Re( eiθ ) and

I = π e−2x cos x dx = Re π e−2x (cos x + i sin x) dx = Re π e−2x eix dx
00 0

The smart thing to do, then, would have been to work out

I = π e−2x+ix dx

0

and take real parts right at the end. This would have halved the effort and would also have doubled
the output should we have wished to evaluate π e−2x sin x dx as well (which is the imaginary part

0

of I ).

Standard Pace Maths Course Complex Variables 13

This would have gone

I= π e−2x +ix dx = − e−(2−i)x π = − e−(2−i)π −1 =

0 2−i 2−i

0

e−2π + 1 = 2+i ( )2 e−2π +1
2−i 22 + 12
( )= e−2π +1 I=
5

Example (ii)
Find

S = cosθ + cos 3θ + cos 5θ + ... + cos (2N + 1)θ

Again, we could substitute for the cosines in terms of complex exponentials, but with an eye to
halving the effort, we consider instead

S = eiθ + ei3θ + ei5θ + ... + ei(2N +1)θ

The aim is to find this first and take the real part to find S.

Expressed like this, we can see that S is a G.P. with first term a = eiθ
and common ratio r = ei2θ . It has as sum

S= ( )a 1− r N +1 = eiθ (1− e2i(N +1)θ )

1− r 1− e2iθ

To find S, which is the real part of this, we could hammer it out by multiplying top and bottom by
the complex conjugate of the denominator and then grinding it down, or we could note that the
denominator can be made into a sine function by taking out the appropriate factor

S = eiθ (1 − e2i( N +1)θ ) = 1 − cos 2( N + 1)θ − i sin 2( N + 1)θ
eiθ (e−iθ − eiθ )
− 2i sinθ

= i (1 − cos 2( N +1)θ ) + sin 2( N + 1)θ

2 sin θ

Hence S = sin 2( N +1)θ

2 sinθ

Again, if we can recast the problem so that we are looking for something complex that has the
answer to our problem as real part, then we win.

Standard Pace Maths Course Complex Variables 14

3.11 Complex Impedance and eiωt

Note: The following is really part of the electrical paper and you will cover it again there. Most
people find it tricky, however, so if you half understand it here, you will be ahead of the game.
There is another chance to master it in the Lent Term when you consider Fourier Series (how to
analyse unsteady functions in terms of their frequency content) or in the Easter Term in Mechanical
Vibrations, or in the ....

We aim to show how complex numbers can simplify calculations in alternating current
circuits. The trick will be to work with something complex, the real part of which is the true
answer. First we need to recap on some basics.

3.11.1 Basics
An alternating voltage V = V0 cosωt across a resistor produces a current

I = V = V0 cosωt I R
RR
(1)
If we express this as the real part of something:

then V is the real part of V = V0eiωt and I is the real part of I = V
R

For a capacitor of capacitance C, the charge Q and the voltage V = V0 cosωt obey:

Q = C V = C V0 cosωt I C

Now current I is the rate of change of charge with time, so differentiating +Q I −Q
gives

dQ = I = C dV V
dt dt

I = - −CV0ω sinω t = CωV0 cos ω t + π
2

We see that, for a capacitor, the voltage lags the current by π/2.

In terms of something complex, again we have cosωt cos π − sinωt sin π
V is the real part of V = V0eiωt 2 2

but this time I is the real part of I = ω CV0eiωt +iπ / 2 = iω C V0eiωt

i.e. I = iω CV (2)

Standard Pace Maths Course Complex Variables 15

cosωt cos π + sinωt sin π
2 2
Across an inductor of inductance L we have:
L
L dI
dt =V = V0 cosωt

I

I = V0 sin ωt = V0 cos ωt − π I
ωL ωL 2

So across an inductor the voltage leads the current by π/2. V
In terms of something complex, again we have

V is the real part of V = V0eiωt

and now I is the real part of I = V0 eiωt −iπ / 2 = − i 1 V0eiωt
ωL ωL

i.e. I = V (3)
iω L

3.11.2 Complex Impedance

Equations (1), (2) and (3) are all a form of Ohm’s Law I = V , provided we use a
Z

complex resistance (more commonly called the impedance when complex)

Component Impedance (Z)
Resistor
Capacitor R

Inductor 1
iω C

iω L

However the really big benefit is that impedances can be combined in the same way as resistances
e.g. impedances in series add. Once we are happy manipulating complex numbers, this enables us to
solve AC circuit problems without explicitly setting up differential equations. Instead we treat them
as resistor networks.

Standard Pace Maths Course Complex Variables 16

Example

Find the voltage across the resistor when there is an applied alternating voltage

V = V0 cosωt applied across the terminals.

IC ∼ 1
I iωC

V = V0cos ωt R VR ∼ R ∼
V= VR

V0eiωt

V~ = I 1 + R and VR = I R
iωC

so that −1 = i2 1 = − i2

VR = R V = 1 V 1 = −i
i
R + 1 1− i
iω C ω CR

To simplify this a bit, define 1 − i = r e−iθ
ω RC

where r = 1 + 1 and θ = tan−1 1
ω2R2C2 ω RC

Hence

V~R = V = V0eiωt +iθ
re−iθ r

Re{V~R } V0
( )VR r
= = cos ωt +θ

Just how much did we save ?

If we had stuck with real variables, then the analysis would have gone something like this.

I = C d (V −VR ) = VR
dt R

so that the equation relating V to VR is

dVR + 1 VR = dV = − V0 ω sin ωt (4)
dt RC dt Complex Variables 17

Standard Pace Maths Course

We can solve this with some effort, or at least we could if we had covered differential equations
(don’t worry – coming next!). Instead we satisfy ourselves that the solution we obtained above
works.

L.H.S. = − V0ω sin (ωt + θ ) + V0 cos (ωt + θ )

r rRC

= − V0 ω 1 sin (ωt + θ ) − 1 cos (ωt + θ )
rω RC
r

= − V0 ω cosθ sin (ωt + θ ) − sinθ cos (ωt + θ )
= − V0 ω sin (ωt + θ −θ ) = − V0 ω sin (ωt )

The saving in effort rises exponentially as the circuit becomes more complicated.

Summary

• Many problems involving trigonometric functions are easier if extended to
complex form, where the answer becomes the real part of the solution to the
extended problem
e−α x cos β x dx → Re e−α x+iβ x dx

cos nθ → Re einθ

• a.c. circuit theory much easier if we use

( )( )V = Re V = Re V0eiωt

( ) ( )( )I = Re I = Re Aeiωt = Re A eiωt +iϕ

• Resistors, capacitors and inductors can then be modelled as resistors with

Resistor → R Capacitor C→ 1 Inductor L→ iωL
iωC

Can now finish Examples 4/3

Standard Pace Maths Course Complex Variables 18