Loci Of Argand Diagrams Flashcards
Loci
A set of points which satisfy some restriction
|z|
The distance from z to the origin
Cartesian form
In terms of x and y
Loci of |z-(x+yi)| = r
- Write in the form |z-(x1 + y1i)| if you haven’t already
- Write z as x + yi |x+yi-x1-y1i| = r
- Group real and imaginary and factor out r |(x-x1) + i(y-y1)|
- Write out as root (x-x1)^2 + (y-y1)^2 = r
- Square both sides (x-x1)^2 + (y-y1)^2 = r^2
- Use that and the centre (x1,y1) to draw the radius
How to find the centre from a mod
Reverse the signs of the values that come after z
Loci of |z-z1| = |z-z2|
Perpendicular bisector of z1 and z2, remembering you need the inverse of what the signs are
Minimising and maximising arg(z) when the circle touches the x-axis
- Draw a tangent to the circle from the origin and a radius at 90 degrees to that
- The x coordinate of the centre is the length
- Draw a line from the origin to the centre to complete the triangle.
- Draw the other tangent from the origin with another radius at 90 degrees
- Fill in the tangent and radius lengths and use tan to find the angle of one triangle then double
Minimising and maximising |z|
- Draw a line from the origin to the centre of the circle
- Use pythagoras to find out the length of that line using the x and y coordinates as the short er sides
- Minimum is length - radius, maximum length + radius
Minimising |z| when |z-z1| = |z-z2|
- You are given z1,z2 and their perpendicular bisector with the y-intercept
- Draw a dotted line from z1 to z2 and a parallel line to that from the origin (length you need)
- The gradient of z1->z2 gives you the gradient of that line, c= 0 as the y-intercept is the origin
- Work out the equation of the perpendicular bisector with gradient the negative reciprocal of |z|’s gradient and c of the y intercept
- Make both equal and solve for x
- Substitute for y
- Use pythagoras for the length
Finding the line which produces an argument at (x1,y1)
- You are given the point and the arg
- Draw a rightward dotted line from the point (this is where the arg comes from)
- Rewrite as (z-(x1 + y1)) with (x1,y1) the point
- Replace z with x + yi (x + yi - x1 - y1) = theta
- Separate x and y by factoring i ((x- x1) + i(y - y1)) = theta
- tan theta = (y-y1)/(x-x1) = arg
- Work out tan theta for the arg
- Multiply that by (x-x1) and rearrange for y =
- Draw that line onto the graph up to the point
- Write the range of x values it applies to
How to sketch |z - a - bi| <= r
Draw circle centre (a,b), radius r
How to draw inequalities on graphs?
> = or <= have solid lines
> or < have dotted lines
What does it mean if you need to sketch?
You don’t need the full equation to draw it
Minimising and maximising arg(z) when the circle doesn’t touch the axis
- Draw a line from the origin to the centre and complete that right angled triangle
- The maximum and minimum will be the two tangents to the circle
- Draw the radius at 90 to the tangent make a right angled triangle with the line from the origin to the centre and the tangent
- Use Pythagoras to find the length of the line from the centre
- Use trig with that length and the radius to find the angle of the triangle from the centre line to the tangent
- Add to the angle of the right angled triangle made with the centre line for the max
- Subtract from that angle if the tangent you have drawn gives the min
Half line equation
arg(z - (a+bi)) = θ
Where a + bi is the endpoint of the half line
