NCEA Level 3 Mathematics 2021
Calculus: Complex Numbers exam
Excellence questions only.
QUESTION ONE
(e) Given that the real part of `(z-2i) / (z-4)` is zero and `z ≠ 4`, prove that the locus of points described by `z` is given by the Cartesian equation `(x-2)^2 + (y-1)^2 = 5`.
The student should be able to recognise: `z` cannot be `4` because that would give a divide by zero; and the equation is the formula for a circle centred at `(2,1)` with a radius `sqrt(5)`.
A big part of problem solving is making an equation and solving it. Basing an equation on a formula is a good way to start, because it provides a pre-baked way to relate values to each other.
We'll work with what we know and write it as an equation, then replace `z` with `x + yi`.
`Re( (z-2i) / (z-4) ) = 0`
`Re( (x + yi - 2i) / (x + yi - 4) ) = 0`
`Re( (x + (y-2)i) / ((x-4) + yi) ) = 0`
If we rationalise the denominator and take only the real terms in the numerator, we can eliminate the need for the Re() function in the equation. There's no need to calculate any of the imaginary terms in the numerator, and there aren't any in the denominator. (That's why we're rationalising the denominator.)
`Re( ((x + (y-2)i) ((x-4) - yi)) / (((x-4) + yi) ((x-4) - yi)) ) = 0`
`(x(x-4) - y(y-2)i^2) / ((x-4)^2 - y^2i^2) = 0`
`(x(x-4) + y(y-2)) / ((x-4)^2 + y^2) = 0`
At this point we can take the numerator and denominator separately.
`x(x-4) + y(y-2) = 0 and (x-4)^2 + y^2 ≠ 0`
`x^2 - 4x + y^2 - 2y = 0`
Since we know the equation we're aiming at has perfect squares in x and y, convert these terms to perfect squares and move any resulting "correction term" constants to the right hand side.
`(x - 2)^2 - 4 + (y-1)^2 - 1 = 0`
`(x - 2)^2 + (y-1)^2 = 5`
Comment: I found this one tricky, with a couple of false starts in my approaches.
QUESTION TWO
(e) If `z` is a complex number and `|z + 16| = 4|z + 1|`, find the value of `|z|`.
Algebra will work fine.
`|z + 16| = 4|z + 1|`
`(z + 16)^2 = 16(z + 1)^2`
`z^2 + 32z + 256 = 16(z^2 + 2z + 1)`
`z^2 + 32z + 256 = 16z^2 + 32z + 16`
The `32z` terms cancel from both sides, so we can most easily solve this equation by moving the `z^2` terms to one side and the constants to the other.
`15z^2 = 240`
`z^2 = 240/15 = 480/30 = 48/3 = 16`
`z = +-sqrt(16)`
`z = +-4`
`|z| = 4`
And there we are. But what is the equation about? What does it all mean?
If we rewrite each side of the equation in the form `z-a` we get each side in a form which shows how a point `z` on the Argand diagram has been moved from the origin.
`|z + 16| = 4|z + 1|`
`|z - (-16)| = 4|z -(-1)|`
The left side therefore involves a point moved 16 left from the origin; the right side has a point moved 1 left from the origin. The modulus signs around each point, `|"this space sells"|`, means we're looking at the distance from the point `z`. The equation is used to fix that distance to a certain value, and this thus defines the radius of a circle.
In this case we get a single circle with its radius being a ratio of the distances from the two points (which works out to be a fixed number). Every point in the locus (ie, on the circle) is four times as far from `z - (-16)` as it is from `z - (-1)`.
Because both of the points have been moved horizontally, we can easily find two points on the circle which intersect with the real (`x`) axis and have the required ratio of distances to (-16, 0) and (-1, 0). These points will be on opposite sides of the circle, giving us a diameter.
`x + 16 = 4(x + 1)`
`x + 16 = 4x + 4`
`3x = 12`
`x = 4`
`x + 16 = -4(x + 1)`
`x + 16 = -4x - 4`
`5x = -20`
`x = -4`
This means the circle defined by the equation is centred at the origin – very convenient – with a radius of 4, so `|z| = 4`.
This understanding of the equation is all very good, but because of the way the question is worded, it's not needed.
Comment: As stated, this question does not seem hard enough to be an excellence question. The assessment schedule breaks the complex number into x and y components before doing the algebra. I can't see any reason to do this when the circle is centred on the real axis.
QUESTION THREE
(e) Solve the equation `z^2 = i(|z|^2 - 4)`.
It's a bit unusual to have a modulus `z` which gets squared. Be careful working with this term.
`z^2 = i(|z|^2 - 4)`
`(x + yi)^2 = i((sqrt(x^2 + y^2))^2 - 4)`
`x^2 + 2xyi + y^2i^2 = i(x^2 + y^2 - 4)`
`x^2 - y^2 = i(x^2 + y^2 - 4) - 2xyi`
Split into separate equations for the real and imaginary components and use them basically as simultaneous equations.
Re: `x^2 - y^2 = 0`
`x^2 = y^2`
`x = +-y`
Im: `x^2 + y^2 - 4 - 2xy = 0`
`x^2 + y^2 - 2xy = 4`
Substitute results from the real components' equation into imaginary components' equation.
`x = y:`
`x^2 + x^2 - 2x^2 = 4`
`0 = 4` is not valid `=> x ≠ y`
`x = -y:`
`x^2 + x^2 + 2x^2 = 4`
`4x^2 = 4`
`x^2 = 1`
`x = 1, y = -1`
`or x = -1, y = 1`
`=> z = 1 - i or z = -1 + i`
Comment: To efficiently answer this question it helps to have strong algebra skills. Methodical working helps to eliminate false solutions.
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