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#### Complex Numbers

• i² = -1
• The general form for all complex numbers is a + bi
• From this, we say:
• Re(a + bi) = a, and
• Im(a + bi) = -a
• NOTE: Re – Real, and Im – Imaginary
• The complex number z and its conjugate z+ are
• z = a + bi, and
• z+ = a – bi
• Arithmetic Operations of Complex Numbers:
• Addition and Subtraction: Add and subtract real, and imaginary parts with each other
• Multiplication: Carry out algebraic expansion, if i² is present, convert it to -1
• Division: Rationalize the denominator by multiplying conjugate pairs
• Equivalence: Equate coefficients
• b² – 4ac is a negative value
• pull out a negative and replace it with i²
• simplify to its general form
• Use sum of two squares
###### Examples
1. Solve z2 + 4z + 13 = 0
• Convert to completed square form: (z + 2)2 + 9 = 0
• Utilize i2 as -1 to make it a difference of two squares: (z + 2)2 – 9i2 = 0
• Proceed with the general difference of two squares method:
• (z + 2 + 3i) (z + 2 – 3i) = 0
• ∴ z = -2 + 3i, and -2 – 3i

• For Square Roots:
###### Examples
1. Find the square roots of 4 + 3i
• We can say that √(4 + 3i) = a + bi
• Square both sides: a2 + b2 + 2abi = 4 + 3i
• Equate the real and imaginary parts: a – b = 4, 2ab = 3
• Solve the equations simultaneously:
• a = 3√2/4 , b = √2/2
• ∴ √(4 + 3i) = 3√2/4 + i√2/2 or –3√2/4 – i√2/2
###### Argand Diagram
• For the complex number z = a + bi:
• Its magnitude is defined as |z| = √(a² + b²)
• Its argument is defined as arg z = tan-1 b/a
• Simply plot the imaginary (y-axis) against the real (x-axis) • Arguments: Always -π < θ < π • The position of z+ is a reflection in the x-axis of z
###### Locus
• |z – w| = r
• The locus of a point z such that |z – w| = r, is a circle with its center at w and with radius r. • arg(z – w) = θ
• The locus of a point z such that arg(z – w) = θ is a ray from w, making an angle θ with positive real axis • The locus of a point z such that |z – w| = |z – v| is the perpendicular bisector of the line joining w and v ###### Examples
1. On a sketch of an Argand diagram, shade the region whose points represent the complex numbers z which satisfy the inequality |z – 3i| ≤ 2. Find the greatest value of arg z for points in this region.
• To find the greatest value of arg z within this region, we must use the tangent at a point on the circle which has the greatest value of θ from the horizontal (red line) • The triangle magnified: • sin α = 2/3
• α = 0.730
• θ = α + π/2
• θ = 0.730 + π/2 = 2.30
2. On a sketch of an Argand diagram, shade the region whose points represent complex numbers satisfying the inequalities |z – 2 + 2i|, arg z ≤ -π1/4, and Re z ≥ 1
• 1. Calculate the greatest possible value of Re z for points lying in the shaded region
• The greatest value for the real part of z would be the one which is furthest right on the Re axis but within the limits of the shaded area. Graphically: • Now using circle and Pythagoras theorems, we can find the value of x:
• x = 2 × cos π/4
• x = √2
• ∴ Greatest value of Re z = 2 + √2
###### Polar Form
• For a complex number z withe magnitude R and argument θ:
• z = R(cos θ + i sin θ) = Re
• ∴ cos θ + i sin θ = e
• Polar Form to General Form:
###### Examples
1. Convert z = 4eπ/4i to the general form
• R = 4, arg z = π/4
• ∴ z = 4(cos π/4 + i sin π/4)
• z = 4 (√2/2 + √2/2i)
• z = 2√2 + (2√2)i

• General Form to Polar Form:
###### Examples
1. Convert z = 2√2 + (2√2)i to polar form
• z = 2√2 + (2√2)i
• R = √((2√2)² + (2√2)²) = 4
• θ = tan-1 2√2/2√2 = π/4
• ∴4(cos π/4 + i sin π/4) = 4eπ/4i
###### Multiplication and Division in Polar Form
• To find the product of two complex numbers in polar form:
• Multiply their magnitudes
• z1z2 = |z1||z2|(arg z1 + arg z2)
###### Examples
1. Find z1z2 in polar form given, z1 = 2(cos π/4 + i sin π/4), z2 = 4(cos π/8 + i sin π/8)
• z1z2 = (2 × 4)(cos π/+ π/8) + i sin (π/+ π/8)
• ∴ z1z2 = 8(cos /8 + i sin /8)

• To find the quotient of two complex numbers in polar form:
• Divide their arguments
• Subtract their magnitudes
• z1/z2 = |z1|/|z2| (arg z1 – arg z2)
###### Examples
1. Find z1/z2 in polar form given, z1 = 2(cos π/4 + i sin π/4), z2 = 4(cos π/8 + i sin π/8)
• z1/z2 = (2/4)(cos π/4 π/8) + i sin (π/4 π/8)
• ∴ z1/z2 = 1/2(cos π/8 + i sin π/8)
###### De Moivre’s Theorem
• zn = Rn(cos nθ + i sin nθ) = Rneinθ
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