Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
The Complex Plane; DeMoivre's Theorem Remember a complex number has a real part and an imaginary part. These are used to plot complex numbers on a complex plane. z x yi z x y 2 Imaginary Axis 2 z x yi z x y Real Axis The magnitude or modulus of z denoted by z is the distance from the origin to the point (x, y). The angle formed from the real axis and a line from the origin to (x, y) is called the argument of z, with requirement that 0 < 2. y tan x 1 modified for quadrant and so that it is between 0 and 2 z x yi We can take complex numbers given as and convert them to polar form. Recall the conversions: z r cos r sin i factor r out r cos i sin y r sin x r cos Imaginary Axis z =r 1 3 x The magnitude or modulus of z is the same as r. y Plot the complex number: z 3 i Real Axis Find the polar form of this number. r 3 1 2 5 5 tan 1 1 z 2 cos i sin 3 6 6 2 4 2 but in Quad II 5 6 The Principal Argument is between - and Imaginary Axis 1 tan but in Quad II 3 1 z =r 1 3 x y Real Axis 5 6 5 5 arg z principal arg 6 6 5 5 z 2 cos i sin 6 6 It is easy to convert from polar to rectangular form because you just work the trig functions and distribute the r through. 5 5 3 1 z 2 cos i sin i 3 i 2 6 6 2 2 1 2 3 2 1 2 3 5 6 If asked to plot the point and it is in polar form, you would plot the angle and radius. Notice that is the same as plotting 3 i Let's try multiplying two complex numbers in polar form together. z1 r1 cos1 i sin 1 z2 r2 cos2 i sin 2 z1 z2 r1 cos 1 i sin 1 r2 cos 2 i sin 2 Look at where andwhere we ended up and r1r2 we cosstarted i sin cos i sin 1 as to what 2 2 see if you can make a1 statement happens to Must FOIL these two complex the r 's and the 's when you multiply numbers. r1r2 cos 1 cos 2 i sin 2 cos 1 i sin 1 cos 2 i 2 sin 1 sin 2 Replace i 2 with -1 and group real terms and then imaginary terms Multiply the Moduli and Add the Arguments r1r2 cos1 cos2 sin 1 sin 2 sin 1 cos2 cos1 sin 2 i use sum formula for cos use sum formula for sin r1r2 cos1 2 i sin 1 2 Let z1 r1 cos 1 i sin 1 and z2 r2 cos 2 i sin 2 be two complex numbers. Then z1 z2 r1r2 cos1 2 i sin 1 2 (This says to multiply two complex numbers in polar form, multiply the moduli and add the arguments) If z2 0, then z1 r1 cos1 2 i sin1 2 z2 r2 (This says to divide two complex numbers in polar form, divide the moduli and subtract the arguments) Let z1 r1 cos 1 i sin 1 and z2 r2 cos 2 i sin 2 be two complex numbers. Then z1 z2 r1r2 cos1 2 i sin 1 2 z1z2 r1r2 cis1 2 If z2 0, then z1 r1 cos1 2 i sin1 2 z2 r2 z1 r1 cis 1 2 z2 r2 If z 4 cos 40 i sin 40 and w 6 cos120 i sin 120 , find : (a) zw (b) z w zw 4 cos 40 i sin 40 6 cos120 i sin120 4 6 cos 40 120 i sin 40 120 multiply the moduli add the arguments (the i sine term will have same argument) 24 cos160 i sin160 24 0.93969 0.34202i 22.55 8.21i If you want the answer in rectangular coordinates simply compute the trig functions and multiply the 24 through. 4 cos 40 i sin 40 z w 6 cos120 i sin 120 4 cos 40 120 i sin 40 120 6 divide the moduli 2 cos 80 3 subtract the arguments i sin 80 In polar form we want an angle between 0° and 180° PRINCIPAL ARGUMENT In rectangular coordinates: 2 0.1736 0.9848i 0.12 0.66i 3 You can repeat this process raising complex numbers to powers. Abraham DeMoivre did this and proved the following theorem: DeMoivre’s Theorem Abraham de Moivre (1667 - 1754) If z rcos i sin is a complex number, then z r cos n i sin n n n where n 1 is a positive integer. This says to raise a complex number to a power, raise the modulus to that power and multiply the argument by that power. This theorem is used to raise complex numbers to powers. It would be a lot of work to find 3 i 3 i 3 i 3 i Instead let's convert to polar form and use DeMoivre's Theorem. r 3 2 1 4 2 2 3 i 4 you would need to FOIL and multiply all of these together and simplify powers of i --- UGH! 1 but in Quad II 5 3 6 tan 1 4 5 5 3 i 2 cos i sin 24 cos 4 5 i sin 4 5 6 6 6 6 4 10 16 cos 3 10 i sin 3 1 3 i 16 2 2 8 8 3i Solve the following over the set of complex numbers: z 1 3 We know that if we cube root both sides we could get 1 but we know that there are 3 roots. So we want the complex cube roots of 1. Using DeMoivre's Theorem with the power being a rational exponent (and therefore meaning a root), we can develop a method for finding complex roots. This leads to the following formula: 2 k 2 k z k r cos i sin n n n n n where k 0, 1, 2, , n 1 Let's try this on our problem. We want the cube roots of 1. We want cube root so our n = 3. Can you convert 1 to polar form? (hint: 1 = 1 + 0i) 1 0 2 2 tan 0 r 1 0 1 1 0 2k 0 2k zk 1cos i sin , for k 0, 1, 2 3 3 3 3 3 Once we build the formula, we use it first with k = 0 and get one root, then with k = 1 to get the second root and finally with k = 2 for last root. We want cube root so use 3 numbers here 2 k 2 k z k r cos i sin n n n n n 0 2k 0 2k zk 1cos i sin 3 3 3 3 3 0 20 z0 1cos 3 3 3 0 20 i sin 3 3 , for k 0, 1, 2 1cos0 i sin 0 1 Here's the root we 0 21 0 21 z1 1cos i sin 3 3 3 3 2 1 3 2 1cos i sin i 2 2 3 3 0 22 0 22 3 z2 1cos i sin 3 3 3 3 already knew. 3 4 4 1cos i sin 3 3 1 3 i 2 2 If you cube any of these numbers you get 1. (Try it and see!) We found the cube roots of 1 were: Let's plot these on the complex plane each line is 1/2 unit 1 3 1 3 1, i, i 2 2 2 2 about 0.9 Notice each of the complex roots has the same magnitude (1). Also the three points are evenly spaced on a circle. This will always be true of complex roots. Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au