3.9 Complex numbers
- Syllabus
- 9709–2028–2029
- Topic
- 3.9
- Level
- A2
For z=x+iy with real x,y:
| Notation | Meaning |
|---|---|
| Rez=x | real part |
| Imz=y | imaginary part (not iy) |
| ∣z∣=x2+y2 | modulus |
| argz=θ | directed angle from positive real axis |
| z∗=x−iy | conjugate |
A non-zero complex number has arguments differing by 2π. Unless specified, use a consistent principal interval such as −π<θ≤π; 0≤θ<2π may also be convenient. arg0 is undefined.
$x+iy=u+iv$ if and only if $x=u$ and $y=v$. Alsozz^*=x^2+y^2=|z|^2.
For z=−1+i3, ∣z∣=2 and a principal argument is 2π/3; its conjugate is −1−i3.
Equality requires both components, not merely equal moduli. Quadrant determines argument; tan−1(y/x) alone is insufficient.
Add complex numbers componentwise, multiply by expanding and replace i² with −1, and divide by multiplying numerator and denominator by the denominator’s conjugate.
After simplifying, write the answer in a+bi form. Equality of complex numbers requires equality of both components.
(1+2i)/(3−i) =[(1+2i)(3+i)]/10=(1+7i)/10.
A complex fraction is not simplified by dividing real parts and imaginary parts separately.
If a polynomial has real coefficients and a+ib (b=0) is a root, then a−ib is also a root. The result is conditional on real coefficients.
Thepairgivestherealquadraticfactor[x-(a+ib)][x-(a-ib)]=(x-a)^2+b^2.
Write the conjugate root immediately, multiply the pair to form a real factor, divide the cubic/quartic polynomial by it, then solve the remaining lower-degree factor and verify the total degree.
If a real cubic has roots $1+i$ and $2$, it also has $1-i$, so a monic polynomial is(x-2)[(x-1)^2+1].
Conjugate pairing is not guaranteed for non-real coefficients. The broad fundamental theorem/root count is not the learning target here.
An Argand diagram has horizontal real axis and vertical imaginary axis. The complex number z=x+iy is represented by point/vector (x,y) from the origin.
| Algebra | Argand meaning |
|---|---|
| Rez=x | horizontal coordinate |
| Imz=y | vertical coordinate |
| ∣z∣ | distance from origin |
| argz | directed angle from positive real axis |
−2+3i is plotted at (−2,3) in quadrant II; its modulus is 13 and its argument must lie in the chosen quadrant-II branch.
Label axes, scale and reference points. Read a plotted point back as real coordinate plus i times imaginary coordinate.
This is a coordinate representation, not a graph y=f(x). Polar multiplication belongs to the next objective.
Writez=r(\cos\theta+i\sin\theta)=re^{i\theta},\qquad r>0.
| Operation | Modulus | Argument |
|---|---|---|
| z1z2 | r1r2 | θ1+θ2 |
| z1/z2 | r1/r2 | θ1−θ2 |
[2e^{i\pi/3}][3e^{-i\pi/6}]=6e^{i\pi/6},\qquad \frac{2e^{i\pi/3}}{3e^{-i\pi/6}}=\frac23e^{i\pi/2}.
Operate moduli and arguments separately, reduce the final argument to the requested interval, and convert to Cartesian form only if asked.
Do not add moduli during multiplication. De Moivre powers and general nth roots are not objectives in this syllabus section and must not be imported.
To solve $z^2=p+iq$, set $z=a+ib$ with real $a,b$:(a+ib)^2=(a^2-b^2)+2abi.Hence $a^2-b^2=p$ and $2ab=q$.
Also $a^2+b^2=|p+iq|=\sqrt{p^2+q^2}$. Add/subtract this with $a^2-b^2=p$ to find $a^2,b^2$, then use $2ab=q$ for signs.
For $z^2=5+12i$:a^2-b^2=5,\quad2ab=12,\quad a^2+b^2=13,so $(a,b)=(3,2)$ or $(-3,-2)$ andz=\pm(3+2i).
Square both answers and show full Cartesian working. The two roots are always opposites for a non-zero complex number.
Do not take square roots of real and imaginary parts separately, and do not report only a principal root.
| Operation | Geometrical effect |
|---|---|
| z↦z∗ | reflect in real axis |
| z↦z+w | translate by vector w |
| z1−z2 | displacement from point z2 to z1 |
| multiply by reiθ | scale distances from origin by r, rotate by θ |
| divide by reiθ | scale by 1/r, rotate by −θ |
Multiplication by i=eiπ/2 rotates every point 90∘ anticlockwise about the origin without changing modulus. Multiplication by −2 scales by 2 and rotates by π.
Addition/subtraction use parallelogram/displacement geometry; multiplication/division use origin-centred scale and rotation. Conjugates keep modulus and negate the argument within branch conventions.
Predict the geometric result, then confirm with Cartesian or polar arithmetic.
A transformation effect is not a locus condition. Powers/general roots are not needed for this objective.
| Condition | Locus |
|---|---|
| ∣z−a∣=r | circle centre a, radius r |
| ∣z−a∣<r / ≤r | interior, boundary excluded/included |
| ∣z−a∣>∣z−b∣ | points closer to b than a: one side of perpendicular bisector |
| ∣z−a∣=∣z−b∣ | perpendicular bisector of segment ab |
| arg(z−a)=α | ray from a at angle α, endpoint a excluded |
Plot reference points first, draw each boundary, decide included/excluded style from equality, shade the correct side/interior/exterior, then intersect all regions.
∣z−(2+i)∣<3 is the open disc centred at (2,1) with radius 3. Adding arg(z−(2+i))=π/4 restricts to points on the corresponding ray that also lie inside the disc.
For equal distances, do not draw a circle: the locus is a straight perpendicular bisector. Test one simple point to choose a side for an inequality.
arg(z−a) is undefined at z=a, and is a direction condition rather than distance. De Moivre powers/roots are unrelated and not required.