Argand Diagrams
The modulus is the length of the line made when: z = x + iy
∣z∣ =
x2 + y 2
The argument is the anticlockwise rotation from the positive real axis (1st
quadrant)
arg(z) = tan−1 ( xy )
Mod arg form is: r(cosθ + i sinθ)
x = rcosθ
y = rsinθ
You can use the fact that cos(θ) = cos(−θ) and −sin(θ) = sin(−θ). To
help with r(cosθ − i sinθ)
For two complex numbers z_1, z_2
∣z1 × z2 ∣ = ∣z1 ∣ × ∣z2 ∣
1∣
∣ zz12 ∣ = ∣z
∣z2 ∣
arg(z1 × z2 ) = z1 + z2
arg( zz12 ) = z1 − z2
So when multiplying complex numbers z1 z2
1. You can expand if in x + iy form but if in mod arg form …
2. Your new modulus will be ∣z1 ∣ × ∣z2 ∣
3. Your new argument will be arg(z1 ) + arg(z2 )
4. Rewrite in mod arg form
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5. i.e. r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 ))
For loci, where z1 = a + ib
∣z∣ is therefore
a2 + b2
(a − 3)2 + b2
∣z − 3∣ =
This is because we have ∣a + ib − 3∣
Group together like terms |(a − 3) + ib∣
|z + 2 − 4i∣ =
2
(a + 2) + (b − 4)
2
For loci where we have z1 which is fixed (i.e. a point we cant move) and z
which is general a points that we can move such that its modulus r, stays
constant. z = x + yi
∣z − z1 ∣ = r, this means….
z − z1 = x + yi − a − bi = (x − a) + (y − b)i
Thus ∣z − z1 ∣ =
2
2
(x − a) + (y − b) = r
So (x − a)2 + (y − b)2 = r2 which is the equation of a circle centre
(a, b) with radius r
note ∣z − z1 | is the difference between the two complex numbers (the
length of this)
For loci (circles), the maximum value of arg(z) occurs when z is a tangent to
the circle.
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Remember z can be any point as it is a general complex number
We can find this angle by remembering that tangents hit the circle at
90degrees to the circle.
And we can make two triangles from this
Both angles made are the same.
Then label what you know (radius, the centre of circle: x,y)
θ1 = tan−1 ( xy ), then double your answer to find the maximum angle
To find the maximum and minimum values of |z| (o.e. what is the longest
and shortest line you can draw from the origin so that you are still touching
the circle)
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|z∣min = ∣z∣max − 2r (where r=radius)
Using Pythagoras and then centre of the circle, you can fins the length to
go the centre of the circle, then add a radius to find ∣z∣max
What does∣z − z1 ∣ = ∣z − z2 ∣ mean ?
Its the distance between the complex numbers z and z1 is the same as the
distance between z and z2
or z is equidistant from z1 and z2
This gives us a perpendicular bisector of the line segment joining z1 and
z2
To find the equation of z .
Write both sides in the form x + iy
Find the modulus of both sides
Square everything
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Cancel (you should have an x2 and y 2 which cancel)
To find the minimum of z
You have to find where two lines intersect
Note the green line pasts through (0,0) and is perpendicular to the red line
What does arg(z − z1 ) = θ mean?
All points where the the angle between the origin and a complex number is
constant this gives a half line. note (0,0) is not included so draw an empty
circle on top of the origin when sketching
so if arg(z) = π6 where z = x + iy
tan(θ) = xy
tan( π6 ) = xy
Thus y = 33 x
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where x< -3 and y>-2
∣z − a − bi∣ = r ⇒ circle, centre (a, b) radius r
arg(z − a − bi) = θ⇒half line centre(a, b) angle θ
The gradient of half lines will be tan(θ)
When a complex number is drawn on a argand diagram, it’s conjugate will be
it’s reflection on the real (x) axis
On an argand diagram multiplying/diving a complex number by i⇒ rotation of
90deg anticlockwise/clockwise
In general when multiplying a complex number z , by a complex number w. z
will undergo …
An anticlockwise rotation of arg(w) about the origin
A scaling of scale factor ∣w∣
when dividing a complex number z , by a complex number w. z will undergo
A clockwise rotation of arg(w) about the origin
A scaling of scale factor ∣ wz ∣
Identity: cos(θ) + isin(θ) = eiθ
so z = reiθ
This exponential form better explains how you can multiply and divide
complex numbers
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Note that all angles repeat themselves every 2π radians, so you may find
hidden solutions if you ±2π to your angle
For Questions you must draw accurate diagrams and pay attention to the word
“touch”
You can use this to help adjust your angles for arctan( xy )
-To deal with questions
You must have arg(z-….) or |z-…|
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