Alg1131W6T1 - Complex arithmetic

Saturday, 14 April 2018

12:14 PM

Machine generated alternative text: A complex number is an expression of the form z a + bi where a and b are real numbers 3 O We add and subtract complex numbers componentwise, so that and (4 — 4i) — (—4+ 3 i) Note: you are being asked for the real and imaginary components separately 2 For multiplication, we just use the usual distributive law, together with the crucial property that i 2 The symbol i represents a new kind of number, whose main property is that i O so that for example It is, of course; useful to remember some basic definitions; if z = a+bi then the real part of z is Re O and the imaginary part of z is Im(z) O Note carefully that the imaginary pan is actually a real number! It follows that for two complex numbers z and w; Re(zw) = and Im(zw) Taking a quotient of two complex numbers is a bit trickier, and involves the notion of the complex conjugate of z = a + bi, defined by a-bi Pleasantly, when we multiply a complex number z = a +bi by its complex conjugate E, we get a real number which might be called the quadrance or modulus squared of z, namely 12 we multiply both numerator and denominator by the complex conjugate of the bottom; namely 5 — 5 i — So to simplify a quotient like 5—5i' Note: enter the complex number a + ib use the Maple syntax a+b*l_ This then gives us 5—5 i 4/5 (5—5 i) (5—5 i)

 

Machine generated alternative text: —9 Given the complex numbers z 8 i and w o fit 6 + i, calculate. 2 ii) w 35+121 iii) w —z 15+91 -(62+391)/37 -(41+631) Note: write the imaginary unit in Maple notation as I (upper case i). For example 3 + 4i would be written as 3 + 4 * I

 

Machine generated alternative text: The origin 0 together with the complex numbers z and w can be completed to form a parallelogram in several ways (larger picture 1m available here). Re Select all possible complex numbers which could be the fourth point of the parallelograrn

 

Machine generated alternative text: 2 Marvin asserts that if both z and w are complex numbers on the unit circle x + y 1, then their product zw will also be Which two of the following complex numbers lie on the unit circle? 5 12 13 13 Write the answer as a subset of {z, w, u, v, t}: Note: the subset {z, w, u} would be written in Maple notation as {z , u} 1 {z,v} 4. u 2i, t' —3 5 4. t 5 18 17 18 The product of z and v above is y = zv So it appears that Marvin might be right Note: make sure to use I (capital i) as the imaginary unit O t] and we may check that In fact Marvin is right: this follows from the important fact that for any two complex numbers z and w v [select all that apply] Q(zw) =

 

Machine generated alternative text: From the approximate positions of the complex numbers z and w on the diagram (larger version available here), identify ii) z — w iii) 2z — w -3 iv) zw

 

Machine generated alternative text: An important use of complex conjugates is the formula for the quadrance, or squared modulus, of a complex number. If z imaginary parts of z. What about z2? We compute that 12 a + bi then we have seen that Q (z) is the sum of the squares of the real and and It follows that the quadrance of z 2 Since this simplifies to [Q (z)] 1m z2 will be the sum of the squares of these: namely 2aö o fit (a2 + b2) we get a remarkably simple way to generate Pythagorean triples using complex numbers! A Pythagorean triple (p, q, r) is a triple of natural 2 2 numbers satisfying p + q For example if z = 2 + i then Q (z) 2 O while z — 3 + 4i. Sowe see that we get the Pythagorean triple {Re(z2), Im(z2), Q(z)} = {3, 4, 5} . Enter the following answers as a set of three numbers: • Using z = 3 + 2i we get the Pythagorean triple: • Using z = 5 + 3i we get the Pythagorean triple: Note: A set in Maple notation is written out in the form {3 , 4, 5}. Challengæ Can you find in this way a Pythagorean triple all of whose elements are bigger than 100, and with no common factors? {119,120,169} O Make sure to enter your answer as a set.

 

 

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