Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 3: Pure Mathematics 3 · official syllabus
COMPLEX NUMBERS LAB
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Four short missions on the Argand plane, division, square roots and loci. Predict the rule, calculate your answer, then test it with a slider. Each completed mission explains the method.
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Mission 1 of 4
Measure on the Argand plane
On an Argand diagram, the horizontal axis is the real part and the vertical axis is the imaginary part. The fixed point is z = 3 + 4i. Set the radius of the circle about the origin so it passes through z: that radius is the modulus |z|.
For z = x + iy, |z| is...
Radius 2.0: compare its square with 3^2 + 4^2 = 25.
Study notes
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Cartesian form and the Argand diagram 3.9
A complex number z=x+iy has real part Re z=x and imaginary part Im z=y, where i^2=-1. Both parts are real numbers. Equality requires equality of both parts separately.
Plot z as (x,y) on an Argand diagram, with real axis horizontal and imaginary axis vertical. Adding complex numbers adds their position vectors; subtraction gives a displacement.
The modulus |z|=sqrt(x^2+y^2) is distance from the origin. For z=3+4i, |z|=5 exactly. Its conjugate z*=3-4i reflects it in the real axis.
Arithmetic and division by a conjugate 3.9
Add or subtract real and imaginary parts separately. For multiplication, expand brackets and replace i^2 by -1.
For division in Cartesian form, multiply numerator and denominator by the denominator's conjugate. The identity zz*=|z|^2 makes the denominator real.
Worked example: (7+i)/(1-2i)=((7+i)(1+2i))/5=(5+15i)/5=1+3i. Check by multiplying the answer by 1-2i: (1+3i)(1-2i)=7+i.
Modulus, argument and polar form 3.9
Write a nonzero number as z=r(cos theta+i sin theta)=r e^(i theta), with r=|z|>0 and theta an argument, measured from the positive real axis. Zero has modulus 0 but no defined argument.
Use the quadrant to choose theta: tan theta=y/x alone cannot distinguish opposite quadrants. For 3+4i, theta=arctan(4/3), approximately 0.9273 radians (4 d.p.). For -3+4i the principal argument is pi-arctan(4/3).
A usual principal-argument convention is -pi < theta <= pi. If the question specifies a different interval, adjust by multiples of 2pi without changing the complex number.
Multiplication and division in polar form 3.9
For multiplication, multiply moduli and add arguments: (r e^(i alpha))(s e^(i beta))=rs e^(i(alpha+beta)). For division by a nonzero number, divide moduli and subtract arguments.
Geometrically, multiplying by r e^(i theta) scales distances from the origin by r and rotates anticlockwise through theta. Multiplying by i is a quarter-turn anticlockwise; dividing by i is a quarter-turn clockwise.
Worked example: 2e^(i pi/6) times 3e^(i pi/3)=6e^(i pi/2)=6i exactly. Arguments may need reducing into the requested interval after adding or subtracting.
Finding both square roots 3.9
Set (a+bi)^2=x+iy and equate real and imaginary parts: a^2-b^2=x and 2ab=y. Also a^2+b^2=|z|, which often makes the algebra easier.
Worked example: z=-5+12i has |z|=13. Add a^2-b^2=-5 and a^2+b^2=13 to get a^2=4. With a=2, 2ab=12 gives b=3; the two roots are 2+3i and -2-3i.
Square each answer to check it. Both signs change together: the two roots are negatives of one another, not generally conjugates. A nonzero complex number has two distinct square roots; zero has only the root zero.
Conjugate roots of real polynomials 3.9
If a polynomial has real coefficients, each non-real root a+bi has its conjugate a-bi as a root too. The real-coefficient condition matters.
Their combined factor is (x-a-bi)(x-a+bi)=(x-a)^2+b^2, which has real coefficients. Use it to divide a cubic or quartic and find the remaining roots.
Worked example: if 1+2i is a root of a real polynomial, 1-2i is another root and x^2-2x+5 is a factor. This fact does not say that every negative or every complex number has its conjugate as a root of an arbitrary polynomial.
Loci on the Argand plane 3.9
|z-a|=r (r>0) is a circle with centre a and radius r; |z-a|<r is its interior without the boundary. With <= the boundary is included.
Worked example: |z-(2+i)|=3 is (x-2)^2+(y-1)^2=9. The centre is (2,1), not the origin.
For distinct a and b, |z-a|=|z-b| is the perpendicular bisector of the segment joining them. An argument condition arg(z-a)=alpha is a ray starting from a in direction alpha, excluding a itself because the argument of zero is undefined.
Shade intersections when two inequalities hold together. Keep excluded boundaries dashed and included boundaries solid, and identify the region rather than drawing unrelated full curves.
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