3 - Complex Numbers (FP1)
What is the easy way of finding | z~1 * z~2 |
| z~1 z~2 | = | z~1 | * | z~2 |
How do you calculate a complex conjugate?
You change the sign of the IMAGINARY PART ONLY
Add 3 + 2i and 7 - 3i
(3 + 2i) + (7 - 3i) 3 + 7 + 2i - 3i 10 - i
Solve (2i)^5
2^5 * i^5 32 * i^2 * i^2 * i 32 * -1 * -1 * i 32i You split it into as many i^2 as possible
Find the square root of 3 + 4i
3 + 4i = (a + bi)^2 Multiply out and move around terms to: 3 + 4i = a^2 - b^2 + i(2ba) So ba = 2 and a^2 - b^2 = 3 a = 2 / b Substitute this into other equation and simplify: b^4 + 3b^2 - 4 = 0 You can factorise quartics just like a quadratic when they only have x^4 and x^2 terms: (b^2 + 4)(b^2 - 1) = 0 b^2 = -4 isn't real so b^2 = 1 So b = 1 or -1 Put this into linear equation to get a = 2 or -2 So the square roots are: 2 + i and - 2 - i
Solve (2 + 3i)(4 + 5i)
8 + 12i + 10i + 15i^2 = 8 + 15(-1) + 22i = -7 + 22i Real part comes first
What is a complex number?
A number made up from a real part and an imaginary part In the form: a + bi
How do you derive the quadratic equation if you are given the complex roots?
Add the roots together to get -b and then reverse the sign to get b (x coefficient) Multiply the roots together to get c (quadratic constant) Exam questions should only ever ask about quadratic with a value of a of 1
How do you add complex numbers?
Use normal algebraic rules and pretend i is like x
Given 3 + i is a root of the below equation Solve the below equation completely 2x^4 - 3x^3 - 39x^2 + 120x - 50 = 0
If 3 + i is a root then 3 - i is also -b = (3 + i) + (3 - i) -b = 6 So b = -6 c = (3 + i)(3 - i) c = 9 + 3i - 3i - i^2 c = 9 - (-1) c = 10 One of the quadratics is x^2 - 6x + 10 Now you divide 2x^4 - 3x^3 - 39x^2 + 120x - 50 By x^2 - 6x + 10 Using algebraic long division This gives 2x^2 + 9x - 5 You solve this using quadratic formula to get: x = -1/2 or 5 All the roots are: 3 + i 3 - i -1/2 5
What are the 2 versions of the notation for magnitude?
If z is the only complex number, r is the magnitude If there are multiple complex numbers (A and B), the magnitude of A would be written |OA|
What is the modulus of a complex number?
It's the magnitude on the Argand diagram
If 3 + 5i = (a + bi)(1 + i) Find a and b
Multiply out: 3 + 5i = a + ai + bi + bi^2 Turn i^2 into -1 and move things without i to the left and imaginary parts to the right: 3 + 5i = a - b + ai + bi Bring i outside bracket on RHS: 3 + 5i = a - b + i(a + b) So we can see a - b = 3 and a + b = 5 Now solve these simultaneous equations
What is the unit of the argument?
The angle in RADIANS
What is the argument of a complex number?
The angle it makes with the positive real axis
What is a complex conjugate?
The conjugate of a + bi is: a - bi
What is the magnitude of a complex number?
The distance from the complex number to the origin on the Argand diagram
What is i?
The notation for sqr(-1) i = sqr(-1)
When else will you have to use this quartic technique?
To solve cubics when you are given 1 root
How do you find the argument from an Argand diagram?
Use tan^-1(opp / adj) Trig
How do you solve any quartic equation if you are given one of the complex roots?
You have the root a + bi so you know that another root must be a - bi You add them to get negative b You multiply them to get c Now you form a quadratic x^2 + bx + c This is one of the quadratic "roots" of the quartic You now use algebraic division to divide the quartic by our quadratic to get the other quadratic "root" You solve this other quadratic to find the other 2 roots of the quartic The 4 roots are: a + bi a - bi The 2 from the second quadratic
How do you convert a complex number from modulus-argument form to a + bi?
You have to draw the diagram and find the height and base of the triangle You could just multiply out but the examiner will want you to draw the diagram and use trig
How do you find the magnitude?
You use Pythagoras
Put into form a + bi: (10 + 5i) / (1 + 2i) How would you approach this question?
You would multiply the top and bottom by the conjugate of the bottom Multiply out the top and bottom Simplify
How can we get from i to a real number?
i^2 = -1
Write sqr(-28) in terms of i
sqr(-28) = sqr(28) * sqr(-1) = 2 * sqr(7) * i = 2i sqr(7) You have to put the i before the square root
Write sqr(-36) in terms of i
sqr(-36) = sqr(36) * sqr(-1) = 6i
What is the symbol for the argument?
theta or Arg z
What are Argand diagrams?
x axis represents real part and y axis represents imaginary part
What is the notation for a complex number?
z
What is modulus-argument form?
z = r(cos(theta) + i sin(theta)) Just put the values of r and theta into this form
What is the notation for the conjugate of a complex number?
z*