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Comlex numbers review Complex Numbers

Complex numbers — Complex Numbers

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Complex Numbers — Complex Numbers

In Year 13 Cambridge 9709, complex numbers extend real numbers by introducing the imaginary unit \(i\), where

\[ i^2=-1. \]

This topic includes imaginary numbers, complex numbers, the complex plane, solving equations, and loci. These ideas are closely linked, so it is important to be confident with algebraic form, Argand diagrams, and standard equation-solving methods.

1. Imaginary numbers

The imaginary unit is defined by

\[ i=\sqrt{-1}, \qquad i^2=-1. \]

Higher powers of \(i\) repeat in a cycle:

\[ i^1=i,\qquad i^2=-1,\qquad i^3=-i,\qquad i^4=1. \]

Then the pattern repeats:

\[ i^5=i,\qquad i^6=-1,\qquad i^7=-i,\qquad i^8=1. \]

2. Complex numbers

Standard form

A complex number is written in the form

\[ z=x+iy, \]

where \(x\) and \(y\) are real numbers.

  • \(x\) is the real part, written \( \operatorname{Re}(z) \)
  • \(y\) is the imaginary part, written \( \operatorname{Im}(z) \)

Examples

\[ 3+2i,\qquad -4+i,\qquad 5-7i. \]

A real number is also a complex number with imaginary part \(0\), and a pure imaginary number has real part \(0\).

Equality of complex numbers

If

\[ a+ib=c+id, \]

then the real parts and imaginary parts must be equal:

\[ a=c,\qquad b=d. \]

Addition and subtraction

\[ (a+ib)+(c+id)=(a+c)+i(b+d) \] \[ (a+ib)-(c+id)=(a-c)+i(b-d). \]

Multiplication

\[ (a+ib)(c+id)=ac+aid+bic+bid^2 \] \[ =(ac-bd)+i(ad+bc). \]

Complex conjugate

The conjugate of \( z=x+iy \) is

\[ \overline{z}=x-iy. \]

Useful results:

\[ z+\overline{z}=2x, \qquad z\overline{z}=x^2+y^2. \]

3. The complex plane

Argand diagram

The complex number \( z=x+iy \) is represented by the point \((x,y)\) on the complex plane.

  • horizontal axis: real axis
  • vertical axis: imaginary axis

Modulus

The modulus of \(z=x+iy\) is its distance from the origin:

\[ |z|=\sqrt{x^2+y^2}. \]

Argument

The argument of \(z\), written \(\arg z\), is the angle between the positive real axis and the line joining the origin to the point representing \(z\).

\[ \tan \theta=\frac{y}{x} \]

but the quadrant must always be checked carefully.

4. Solving equations

Linear equations

Solve in the same way as normal algebra, then simplify into the form \(a+ib\).

Quadratic equations

Use factorisation or the quadratic formula. If the discriminant is negative, the roots are complex.

\[ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}. \]

Using conjugates in division

To simplify \( \dfrac{a+ib}{c+id}, \) multiply top and bottom by the conjugate of the denominator:

\[ \frac{a+ib}{c+id}\times \frac{c-id}{c-id}. \]

This makes the denominator real because \[ (c+id)(c-id)=c^2+d^2. \]

5. Loci in the complex plane

Distance from the origin

The locus

\[ |z|=r \]

is a circle with centre \(O(0,0)\) and radius \(r\).

Distance from a fixed point

The locus

\[ |z-a|=r \]

is a circle with centre at the point representing \(a\) and radius \(r\).

Equal distances

The locus

\[ |z-a|=|z-b| \]

is the perpendicular bisector of the line segment joining the points representing \(a\) and \(b\).

Worked examples

Example 1: Simplify powers of \(i\)

Simplify \(i^{27}\).

Solution

Powers of \(i\) repeat every \(4\):

\[ 27=24+3. \]
\[ i^{27}=i^{24}i^3=(i^4)^6 i^3=1^6(-i)=-i. \]
\[ i^{27}=-i. \]

Example 2: Multiply complex numbers

Simplify \((3+2i)(4-i)\).

Solution

\[ (3+2i)(4-i)=12-3i+8i-2i^2 \] \[ =12+5i+2 \] \[ =14+5i. \]
\[ (3+2i)(4-i)=14+5i. \]

Example 3: Divide complex numbers

Express \( \dfrac{3+4i}{1-2i} \) in the form \(a+ib\).

Solution

\[ \frac{3+4i}{1-2i}\times \frac{1+2i}{1+2i} = \frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}. \]
\[ (3+4i)(1+2i)=3+6i+4i+8i^2=3+10i-8=-5+10i \] \[ (1-2i)(1+2i)=1+4=5. \]
\[ \frac{3+4i}{1-2i}=\frac{-5+10i}{5}=-1+2i. \]
\[ \frac{3+4i}{1-2i}=-1+2i. \]

Example 4: Solve a quadratic equation

Solve

\[ z^2-6z+13=0. \]

Solution

\[ z=\frac{6\pm \sqrt{36-52}}{2} = \frac{6\pm \sqrt{-16}}{2} \] \[ = \frac{6\pm 4i}{2} = 3\pm 2i. \]
The roots are \[ z=3+2i \qquad \text{and} \qquad z=3-2i. \]

Example 5: Modulus and argument

For \(z=-1+\sqrt{3}i\), find \( |z| \) and \( \arg z \).

Solution

Modulus:

\[ |z|=\sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{1+3}=2. \]

Since the point lies in the second quadrant,

\[ \arg z=\frac{2\pi}{3}. \]
\[ |z|=2, \qquad \arg z=\frac{2\pi}{3}. \]

Example 6: Interpret a locus

Describe the locus

\[ |z-(2+i)|=3. \]

Solution

The point \(2+i\) represents \((2,1)\) on the Argand diagram.

The locus is a circle with centre \((2,1)\) and radius \(3\).

Common mistakes and exam tips

Common mistakes

  • Forgetting that \(i^2=-1\), not \(1\).
  • Not collecting real and imaginary parts separately.
  • Making sign errors when multiplying complex numbers.
  • Dividing without multiplying by the conjugate.
  • Finding the argument from \(\tan^{-1}(y/x)\) without checking the quadrant.
  • Confusing \(|z|\) with \(\arg z\).
  • In loci questions, not interpreting the modulus as a distance.

Exam tips

  • Always write the final answer in the form \(a+ib\) unless another form is requested.
  • Use the cycle of powers of \(i\) to simplify quickly.
  • For division, write the conjugate step clearly.
  • In Argand diagram questions, sketching the point often helps with modulus, argument, and loci.
  • For a locus such as \(|z-a|=r\), think immediately: distance from a fixed point.

Summary

\[ i^2=-1 \] \[ z=x+iy \] \[ \overline{z}=x-iy \] \[ |z|=\sqrt{x^2+y^2} \] \[ (a+ib)(a-ib)=a^2+b^2 \]

Complex numbers combine algebra and geometry. You should be able to simplify expressions, solve equations, interpret points on the complex plane, and describe loci accurately.

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