Study Notes: Pure Mathematics 3 (Paper 3)
Chapter 3.9: Complex Numbers
Welcome to the world of Complex Numbers! Don’t worry if this sounds scary—it’s just a clever way to solve equations that normal numbers can't handle. This chapter connects algebra, geometry, and trigonometry, making it one of the most powerful and beautiful tools in advanced mathematics. You'll need this foundation for areas like electrical engineering, quantum mechanics, and fluid dynamics.
Our journey begins with the concept that makes complex numbers possible: the imaginary unit.
1. The Basics: Introducing \(i\) and Cartesian Form
Before complex numbers, equations like \(x^2 + 1 = 0\) had no solution in the real number system. Complex numbers solve this problem by inventing a new number.
The Imaginary Unit \(i\)
The Imaginary Unit, denoted \(i\), is defined such that:
$$i^2 = -1$$
$$i = \sqrt{-1}$$
From this, the powers of \(i\) cycle:
$$i^1 = i$$
$$i^2 = -1$$
$$i^3 = -i$$
$$i^4 = 1$$
(This cycle repeats every four powers. This is a great memory aid!)
Definition and Key Terminology (Cartesian Form)
A complex number, typically denoted by \(z\), is written in Cartesian form as:
$$z = x + iy$$where \(x\) and \(y\) are real numbers.
• The Real Part of \(z\) is \(x\). (Notation: \(\text{Re}z = x\))
• The Imaginary Part of \(z\) is \(y\). (Notation: \(\text{Im}z = y\). Note: The imaginary part is the real number \(y\), not \(iy\)!)
• Two complex numbers \(z_1\) and \(z_2\) are equal if and only if (\(z_1 = z_2\) iff) their real parts are equal and their imaginary parts are equal.
1.1 Arithmetic Operations in Cartesian Form
Performing operations on complex numbers in Cartesian form (\(x + iy\)) is often similar to manipulating algebraic expressions involving \(x\) and \(y\), but remembering that \(i^2 = -1\) is key.
A. Addition and Subtraction: Add/subtract the real parts and the imaginary parts separately.
Example: If \(z_1 = 3 + 4i\) and \(z_2 = 1 - i\).
$$z_1 + z_2 = (3+1) + (4i - i) = 4 + 3i$$
B. Multiplication: Multiply out like binomials, remembering to substitute \(-1\) for \(i^2\).
Example: If \(z_1 = 3 + 4i\) and \(z_2 = 1 - i\).
$$z_1 z_2 = (3)(1) + (3)(-i) + (4i)(1) + (4i)(-i)$$
$$z_1 z_2 = 3 - 3i + 4i - 4i^2$$
$$z_1 z_2 = 3 + i - 4(-1) = 7 + i$$
C. Division (The Conjugate Trick): Division requires a special trick to rationalise the denominator.
The Complex Conjugate (\(z^*\))
The complex conjugate of \(z = x + iy\) is \(z^* = x - iy\). You simply change the sign of the imaginary part.
• Did you know? When you multiply a complex number by its conjugate, the result is always a real number (a great way to eliminate \(i\) from the denominator!).
$$z z^* = (x + iy)(x - iy) = x^2 - (iy)^2 = x^2 - i^2 y^2 = x^2 + y^2$$
Division Step-by-Step
To divide \(z_1\) by \(z_2\), multiply the numerator and the denominator by the conjugate of the denominator, \(z_2^*\):
$$ \frac{z_1}{z_2} = \frac{z_1}{z_2} \times \frac{z_2^*}{z_2^*}$$Example: Simplify \(\frac{3 + 4i}{1 - i}\). (The conjugate of \(1 - i\) is \(1 + i\)).
$$ \frac{3 + 4i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(3+4i)(1+i)}{(1-i)(1+i)}$$
$$ \text{Numerator: } 3 + 3i + 4i + 4i^2 = 3 + 7i - 4 = -1 + 7i$$
$$ \text{Denominator: } 1^2 + 1^2 = 2$$
$$ \text{Result: } \frac{-1 + 7i}{2} = -\frac{1}{2} + \frac{7}{2}i $$
1.2 Polynomial Equations and Conjugate Roots
This is a fundamental result for solving polynomial equations:
• If a polynomial equation has real coefficients, then any non-real (complex) roots must occur in conjugate pairs.
Analogy: If you have a matching set of earrings, and one is complex (\(a+ib\)), the other must be its conjugate (\(a-ib\)).
Implication: If you are told that \(z=1+2i\) is a root of a cubic equation with real coefficients, you immediately know that \(z^* = 1-2i\) must also be a root.
Key Takeaway for Section 1: Treat complex arithmetic like algebra, but use \(i^2 = -1\). For division, always use the complex conjugate of the denominator.
2. Geometric Representation: The Argand Diagram
Just as we use an \(xy\)-plane for real numbers, we use the Argand diagram to plot complex numbers. This allows us to visualize complex operations geometrically.
• The horizontal axis is the Real Axis (\(x\)).
• The vertical axis is the Imaginary Axis (\(y\)).
The complex number \(z = x + iy\) is plotted as the point \((x, y)\) or represented by the position vector from the origin to that point.
2.1 Modulus and Argument (Polar Coordinates)
In the Argand diagram, the position of \(z\) can be described using distance and direction, which are the modulus and the argument.
A. Modulus (\(|z|\) or \(r\)):
The modulus is the distance from the origin to the point \(z\). It is always non-negative.
Analogy: This is simply Pythagoras' Theorem!
B. Argument (\(\arg z\) or \(\theta\)):
The argument is the angle between the positive real axis and the line segment connecting the origin to \(z\).
• The argument is usually measured in radians.
• The principal argument is typically defined in the interval \(-\pi < \theta \leq \pi\) (or \(0 \leq \theta < 2\pi\) if specified by the question).
Calculating the Argument:
1. Find the reference angle \(\alpha = \arctan\left|\frac{y}{x}\right|\) (always use the positive values of \(x\) and \(y\)).
2. Determine the quadrant of \(z = x + iy\):
• Quadrant 1 (\(x>0, y>0\)): \(\theta = \alpha\)
• Quadrant 2 (\(x<0, y>0\)): \(\theta = \pi - \alpha\)
• Quadrant 3 (\(x<0, y<0\)): \(\theta = -(\pi - \alpha)\) or \(\theta = \alpha - \pi\)
• Quadrant 4 (\(x>0, y<0\)): \(\theta = -\alpha\)
Common Mistake: The function \(\tan^{-1}\) on a calculator only returns results in Q1 or Q4. You must check the signs of \(x\) and \(y\) to place the complex number in the correct quadrant and adjust the angle accordingly!
Key Takeaway for Section 2: Modulus is distance, calculated using Pythagoras. Argument is angle, calculated using arctan, adjusted for the correct quadrant based on the signs of \(x\) and \(y\).
3. Polar and Exponential Forms (\(re^{i\theta}\))
Once we have the modulus \(r\) and argument \(\theta\), we can express \(z\) in Polar Form:
$$z = r(\cos\theta + i\sin\theta)$$And the highly convenient Exponential Form:
$$z = re^{i\theta}$$These forms are essential for performing multiplication and division quickly, especially when dealing with high powers (though high powers beyond simple squares are usually deferred to Further Maths).
3.1 Multiplication and Division in Polar Form
If \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\):
A. Multiplication: Multiply moduli, add arguments.
$$z_1 z_2 = (r_1 r_2) e^{i(\theta_1 + \theta_2)}$$• Modulus: \(|z_1 z_2| = |z_1| |z_2| = r_1 r_2\)
• Argument: \(\arg(z_1 z_2) = \arg z_1 + \arg z_2 = \theta_1 + \theta_2\)
Geometric Effect: Multiplying by \(z_2\) stretches \(z_1\) by a factor of \(r_2\) and rotates it anticlockwise by angle \(\theta_2\).
B. Division: Divide moduli, subtract arguments.
$$\frac{z_1}{z_2} = \left(\frac{r_1}{r_2}\right) e^{i(\theta_1 - \theta_2)}$$• Modulus: \(\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} = \frac{r_1}{r_2}\)
• Argument: \(\arg\left(\frac{z_1}{z_2}\right) = \arg z_1 - \arg z_2 = \theta_1 - \theta_2\)
Key Takeaway for Section 3: Polar form simplifies multiplication (multiply \(r\)'s, add \(\theta\)'s) and division (divide \(r\)'s, subtract \(\theta\)'s).
4. Advanced Operations and Loci
4.1 Finding the Two Square Roots of a Complex Number
Finding \(\sqrt{w}\) where \(w = a + ib\) results in two complex roots, \(z = x + iy\).
Process (Using Cartesian Form):
1. Assume the square root is \(z = x + iy\).
2. Square both sides: \((x + iy)^2 = a + ib\).
$$x^2 + 2ixy + (iy)^2 = a + ib$$
$$x^2 - y^2 + i(2xy) = a + ib$$
3. Equate the real parts and imaginary parts to form simultaneous equations:
$$ \text{Real part: } x^2 - y^2 = a \quad \quad \text{(Equation 1)}$$
$$ \text{Imaginary part: } 2xy = b \quad \quad \text{(Equation 2)}$$
4. Use Equation 2 to express \(y\) in terms of \(x\) (or vice versa), and substitute into Equation 1.
5. Solve for \(x\) (this usually results in a quadratic equation in \(x^2\)). Find the corresponding values of \(y\).
6. Since \(b\) can be positive or negative, check the signs using \(2xy = b\). If \(b\) is positive, \(x\) and \(y\) have the same sign. If \(b\) is negative, they have opposite signs.
Note: Full details of this working must be shown in the examination.
4.2 Loci in the Argand Diagram (Geometric Effects)
A locus (plural: loci) is a set of points that satisfy a geometric condition. In complex numbers, simple equations/inequalities define specific shapes in the Argand diagram.
Let \(z\) be a general point, and \(a\) and \(b\) be fixed complex numbers corresponding to points \(A\) and \(B\).
Case 1: Distance from a fixed point (\(|z - a| = k\))
• The term \(|z - a|\) represents the distance between the point \(z\) and the fixed point \(a\).
• Locus: A Circle with centre \(A\) and radius \(k\).
$$|z - (1 + 2i)| = 3$$ Interpretation: All points \(z\) that are exactly 3 units away from the point \((1, 2)\). (If it were \(< k\), it would be the interior of the circle.)
Case 2: Equidistant from two fixed points (\(|z - a| = |z - b|\))
• The distance from \(z\) to \(a\) is equal to the distance from \(z\) to \(b\).
• Locus: The Perpendicular Bisector of the line segment \(AB\).
$$|z - 4| = |z - 2i|$$ Interpretation: All points \(z\) equidistant from \(A(4, 0)\) and \(B(0, 2)\).
Case 3: Argument is constant (\(\arg(z - a) = \alpha\))
• This represents the angle that the vector \(\vec{AZ}\) makes with the positive real axis.
• Locus: A Half-Line or Ray originating from the fixed point \(A\).
$$\arg(z - (-2)) = \frac{\pi}{4}$$ Interpretation: A ray starting at \(A(-2, 0)\) and extending at an angle of \(45^\circ\) (\(\pi/4\) radians). The point \(A\) itself is not included in the locus, as the argument is undefined at the origin of the vector.
C. Geometric Effects of Conjugation and Arithmetic:
• Conjugation (\(z \to z^*\)): This is a reflection in the real axis.
• Addition/Subtraction (\(z_1 \pm z_2\)): This is vector addition/subtraction (parallelogram rule).
• Multiplication/Division: These cause rotation and scaling (see Polar Form section).
Key Takeaway for Section 4: Complex number equations often define basic geometric shapes. Memorize the three main locus types (Circle, Perpendicular Bisector, Ray).