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Complex Numbers i.

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Presentation on theme: "Complex Numbers i."— Presentation transcript:

1 Complex Numbers i

2 Given that x² = 2 Or that if ax² + bx + c = 0 Then x = ±√2 Then x = -b ± √b² - 4ac 2a How do we solve x² = -2 or a quadratic with b² - 4ac < 0 ? We can introduce a new number called i which has the property that i² = -1 ie i = √-1. Hence we can solve the equation x² = -2 x = ±√-2 x = ±√2 √-1 x = ±√2 i

3 Example Solve x² + 2x + 2 = 0 x = -b ± √b² - 4ac 2a x = -2 ± √2² - 4x1x2 2x1 x = -2 ± √4 - 8 2 x = -2 ± √-4 2 x = -2 ± √4 √-1 2 x = -2 ± √4 i 2 x = -2 ± 2 i 2 x = -1 ± i

4 i is an imaginary number.
The set of complex numbers, C, are the family of numbers of the form, a + i b, where a and b are real numbers. If z is a complex numbers, C, where z = a + i b, Then a is the real part of z, written Re z. B is the imaginary part of z, written Im z.

5 The Argand Diagram. The Argand Diagram gives a geometric representation of complex numbers as points in the plane R². ie z = x + i y can be expressed by the point (x,y). Im ● (-4,2) = i ● (1,1) = 1 + i Re ● (3,-1) = 3 - i (-2,-2) = i

6 Arithmetic Operations
Addition z1 = a + i b , z2 = c + i d z1 + z2 = (a + c) + i (b + d) e.g. (2 + 3i) + (5 – 4i) = 7 – i Subtraction z1 = a + i b , z2 = c + i d z1 - z2 = (a - c) + i (b - d) e.g. (9 + 2i) - (5 – 6i) = 4 + 8i

7 Arithmetic Operations
Multiplication z1 = a + i b , z2 = c + i d z1z2 = (a + ib)(c + id) = ac + iad + ibc + i²bd e.g. (3 + 2i)(5 - i) = i + 10i - 2i² = i – 2(-1) = i

8 Complex Conjugates If z = a + i b then the complex conjugate of z is defined to be a – ib and denoted z Division e.g. (4 + 2i) ÷ (2 + 3i) = (4 + 2i) (2 + 3i) (2 - 3i) = i + 4i – 6i² 4 - 6i + 6i – 9i² x = 14 – 8i 13 = 14 – 8i

9 The Argand Diagram. Im The distance of any complex number z from the origin is √(x²+ y²). This is called the modulus of z and is written |z|. ● (x,y) y θ Re x The angle θ is the angle of rotation from the real axis to OZ and is called the argument of z or arg(z). tan θ = Y x -π < θ < π Y x θ = tan


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