Complex Numbers. 1 Introduction. 2 Imaginary Number. December 11, Multiplication of Imaginary Number

Transcription

1 Complex Numbers December, 206 Introduction 2 Imaginary Number In your study of mathematics, you may have noticed that some quadratic equations do not have any real number solutions. For example, try as you may, you will never be able to find a real number solution to the equation x 2 =. This is because it is impossible to square a real number and get a negative number! However, a solution to the equation x 2 = does exist in a new number system called the complex number system. Mathematician introduced an imaginary number i = to settle this problem. By using this, any polynomial can be solved. Engineers use j sometimes instead of i. In programming language, both i and j are defined as imaginary number. Write these in terms of imaginary number Multiplication of Imaginary Number i = i 2 = i 3 = i 4 = i 5 =

2 3 Complex Number Complex does not mean complicated. Complex in this context means together, for which we write imaginary number with real number. We write complex number in the standard form a + bi where a and b is a real number. There are other forms such as Euler form and polar form which we will see later. But for instance when want to find the roots, we usually write it conventionally as z = x + iy. See that we used letter z since we have used x and y to represent real number. Also, another way is to write complex number separately such as Re(z) and Im(z) that represent real and imaginary part respectively. Example Desribe the real number in standard form and Re and Im part. z = i 2. z = 3 + 4i 3. z = 5 4. z = 2 i Addition and substraction of Complex Number Example 2. (9 2i) + (3 6i) 2. z = 6 + 5i z 2 = 3 + 2i z z 2 = z + z 2 = 5 Multiplication of Complex Number Example 3. 7i(4 3i) 2. ( + 9i 2 3. (3 5i)(5 + 3i) 4. (4 3i)(2 + 3i)(2 + 2i) 2

3 6 Complex Conjugate The conjugate for z = x + iy is given by z = x iy. Notice that only the imaginary part is the inverse of the original value. Also, it should be noted that the relationship is vice versa, which means the conjugate for z is z. Conjugate is useful when dealing with division of complex number, to change it into the standard form a + bi Example 4 If z = x + iy, the conjugate of z, written as z is z = x iy. Find the conjugate of the following. 2 + i i i i 5. 2 i 6. cos θ + i sin θ 7. α + βi Multiplication of z and its conjugate.. ( + 3i)(2 3i) 2. z = 3 + 2i Find z z 7 Division We see that if we multiply complex number with its conjugate, we will obtain a real number. Thus we can change the division form of complex number to obtain the standard form a + bi. It is done by multiplying the conjugate of the denominator. Example 5 Write the given complex number in standard form of a + bi i 3 + i (2 i)(4 + i) 4 i 3 i (4 + i) 2 3

4 4. 2 i 4i 5. Given z = 2i. Find the values of x and y if z + = x + iy. z 6. Given z = + 2i, w = 2 i, u = 4 3i. Simplify the following expression in the form of a + bi. z + 3 w 2u 2w + ū + z 8 Polar Form Before we learn how to write complex number in polar form, it is helpful to plot the complex number first into the graph. The reason is to find the angle of the point, correspond to the complex number. Just like in polar coordinate before, we can plot complex number z = x + iy into complex plane, known as Argand diagram. Two write complex number in polar form, two things we need. z = x + iy, then the modulus of z denoted by z is given by z = x 2 + y 2. Modulus : If Argument : If z = x + iy, then the argument of z, denoted by arg(z) is arg(z) = tan - y. Note that θ can be in degree or radian. x We usually use r to define z and θ to represent arg(z). The tricky part in writing polar form of complex number is to find the argument of z.when we plot the point of z into Argand diagram, we can actually estimate the value of θ based on which quadrants it s located. But when we used calculator to find θ, it gives us the value of θ defined as in chapter 2, where it didn t really refer to which quadrant the point is located. Another thing to consider is, we usually want arg(z) to be the principal argument, or principal value. interval π < θ π. Step to write z = x + iy in polar form z = r cos θ + i sin θ.. Find the modulus z. Principal argument, denoted by Arg(z) lies in the 2. Plot the x and y into Argand diagram. Take notes which quadrant it is located. 3. Input θ = tan - y into your calculator. If the value of x or y has negative sign, x dont bother about it, just take the positive value of x and y to find θ. Again, θ can be in degrees or radian, no restriction. 4

5 4. Once we have θ from calculator, compared the value with the point in quadrant. If the point z = x + iy is located in first quadrant, just take the value you got from calculator directly. If it is in second quadrant then arg(z) = 80 θ. If the point z is in third quadrant, arg(z) = 80 + θ. If the point z is in fourth quadrant, arg(z) = 360 θ. 5. Now that we have arg(z) we can change it into principal argument. If arg(z) lies in between π and π then it is considered the principle argument already. But if it is not, then we can find principal value Arg(z) = arg(z) 360 where arg(z) is obtained from the previous step. 6. Finally we can write the complex number in the polar form of z = r(cos θ + i sin θ). Example 6 Write the following complex numbers in polar form i i i i i 3 + i Multiplication and division of complex number in polar form Example 6 Solve the following. Given p = 3 i and q = + 3i. Find i. pq p ii. q iii. p 2 iv. p 2 q 2. Find (cos θ + i sin θ) 2 3. Find (cos θ + i sin θ) 3 5

6 9 De Moivre s Theorem De Moivre stated that for any rational value of n, value of (cos θ + i sin θ) n is given by (cos θ + i sin θ) n = cos nθ + i sin nθ Example 7 Find the following using De Moivre s Theorem. (cos θ + i sin θ) 5 2. (cos θ + i sin θ) 4 3. (cos θ + i sin θ) 4 4. ( + i) 6 5. ( i) 4 6. (2 2i) 3 Example 8 Solve the following by utilising De Moivre s theorem. Express cos 5θ cos θ and sin 5θ sin θ in terms of cos θ. 2. Express cos 4θ and sin 4θ in terms of sin θ. 3. Express sin 5 θ in terms of sines of multiples of θ. 0 Application : Finding roots Remember quadratic formula? How many roots does it have? For example, how many roots x 2 6x + 9 = 0 have? How about x 4 5x 3 = 0? Now the idea is to use all that we have learnt this far to find roots. Needless to say, we will always obtain the roots in complex number. Example 9 Solve the following. Solve 4z 2 + (20 2i)z = 34i 3 2. The cube roots of i 3. The fourth roots of 3 4i 4. z 6 = 0 6

7 5. Final Exam 205/206 Question. a) Given a complex number u = 2 + 3i. i. Determine z = u i in the form a + ib. ii. Express z in polar form. iii. Solve w 3 = z and sketch the roots on a single Argand Diagram. b) Use de Moivres theorem to show that sin 3θ = 3 sin θ 4 sin 3 θ. Hence, obtain all solutions of x for the following equation: 4x 3 3x + = 0. 7