Complex Numbers Section 1: Introduction to Complex Numbers Notes nd Exmples These notes contin subsections on The number system Adding nd subtrcting complex numbers Multiplying complex numbers Complex conjugtes Equtions with complex roots Equting rel nd imginry prts The number system In your lerning of mthemtics, you hve come cross different types of number t different stges. Ech time you were introduced to new set of numbers, this llowed you to solve wider rnge of problems. The first numbers tht you cme cross were the counting numbers (nturl numbers). These llowed you to solve equtions like x + = 5. Lter you would meet negtive numbers, which llowed you to solve equtions like x + 5 =, nd rtionl numbers, which ment you could solve equtions like x = 5. When irrtionl numbers were included, you could solve equtions like x² =. However, there re still equtions which you cnnot solve, such s x² = -4. You know tht there re no rel numbers which stisfy this eqution. However, this eqution, nd others like it, cn be solved using imginry numbers, which re bsed on the number i, which is defined s 1. The digrm below shows the reltionships between different types of number. rel numbers rtionl numbers Irrtionl numbers positive integers negtive integers zero MEI, 17/03/06 1/7
This type of digrm is clled Venn digrm (you my hve met these before if you hve studied ny Sttistics) nd it shows the reltionships between sets, in this cse sets of numbers. This digrm dels with the rel numbers, which include ll numbers which you hve come cross until now. Notice tht the positive nd negtive integers (whole numbers) re subsets of the rtionl numbers. This mens tht ll integers re lso rtionl numbers, but there re other rtionl numbers which re not integers, such s 3 or 7 11. Similrly, ll rtionl numbers re rel numbers, but there re other rel numbers which re not rtionl, such s 3 nd π. In this chpter you will see tht the rel numbers re lso subset of lrger set clled the complex numbers. You will be looking t numbers which lie outside the set of rel numbers. Complex numbers involve both rel nd imginry numbers. Adding nd subtrcting complex numbers To dd two complex numbers, you need to dd the rel prts nd dd the imginry prts. Similrly, to subtrct one complex number from nother, del with the rel nd imginry prts seprtely. Exmple 1 The complex numbers z nd w re given by z = 3 + i w = 1 4i Find: (i) z + w (ii) z w (iii) w z (i) z + w = (3 + i) + (1 4i) = (3 + 1) + (i 4i) = 4 i (ii) z w = (3 + i) (1 4i) = (3 1) + (i + 4i) = + 6i Add the rel prts nd dd the imginry prts Subtrct the rel prts nd subtrct the imginry prts (iii) w z = (1 4i) (3 + i) = (1 3) + (-4i i) = - 6i For prctice in exmples like the one bove, try the interctive resource Addition nd subtrction of complex numbers. MEI, 17/03/06 /7
Multiplying complex numbers Multipliction of two complex numbers is similr to multiplying out pir of brckets. Ech term in the first brcket must be multiplied by ech term in the second brcket. You cn then simplify, remembering tht i² = -1. Exmple Find (i) (3 + 4i)( + i) (ii) ( i)(3 + i) (iii) ( + 3i)( 3i) (i) (3 + 4i)( + i) = 6 + 3i + 8i + 4i² = 6 + 11i 4 = + 11i Using i² = -1 Multiply out the brckets (ii) (4 i)(3 + i) = 1 + 8i 3i i² = 1 + 5i + = 14 + 5i (iii) ( + 3i)( 3i) = 4 6i + 6i 9i² = 4 + 9 = 13 For prctice in exmples like the one bove, try the interctive resource Multipliction of complex numbers. Complex conjugtes In prt (iii) of Exmple, the result of multiplying two complex numbers is rel number. This is lwys the cse when the complex number + bi is multiplied by the complex number bi. The complex number bi is clled the complex conjugte of + bi. For ny complex number z, the complex conjugte is written s z or z*. For prctice in complex conjugtes, try the interctive resource Conjugte of complex number. Equtions with complex roots When you first lerned to solve qudrtic equtions using the qudrtic formul, you found tht some qudrtic eqution hd no rel solutions. However, using complex numbers you cn find solve ll qudrtic equtions. MEI, 17/03/06 3/7
Exmple 3 Solve the qudrtic eqution x² + 6x + 13 = 0 AQA Further Pure 1 Using the qudrtic formul with = 1, b = 6, c = 13 b± b 4c x = 6± 36 4 1 13 = 1 6± 16 = 6± 4i = = 3± i The solutions of the eqution re x = -3 + i nd x = -3 i Notice tht the qudrtic eqution in Exmple 3 hs two complex solutions which re pir of complex conjugtes. All qudrtic equtions with rel coefficients hve two solutions: either two rel solutions (which could be repeted solution) or two complex solutions which re pir of complex conjugtes. The next exmple shows how you cn find qudrtic eqution with roots t prticulr complex vlues. A qudrtic eqution with roots t x = nd x = b cn be written s (x )(x b) = 0, nd this lso pplies to situtions where the roots re complex numbers. Alterntively, you cn consider the sum nd the products of the roots. Both pproches re shown in Exmple 4. Exmple 4 Find the qudrtic eqution which hs roots t x = 4 + i nd x = 4 i. 1 (x (4 + i))(x (4 i)) = 0 (x 4 i)(x 4 + i) = 0 (x 4)² (i)² = 0 x² 8x + 16 + 4 = 0 x² 8x + 0 = 0 The two middle terms cncel out since this expression is of the form (x )(x + ) b The sum of the roots is 4 + i + 4 i = 8 = 8 c The product of the roots is (4 + i)(4 i) = 16 + 4 = 0 = 0 Tking = 1 gives b = -8 nd c = 0 The qudrtic eqution is x² 8x + 0 = 0. MEI, 17/03/06 4/7
The Flsh resource Working with complex numbers tests you on multipliction, complex conjugtes nd equtions with complex roots. You cn lso try the Complex numbers Hexgonl puzzle. Using complex numbers, you hve seen tht ll qudrtic equtions hve two solutions. In similr wy, ll cubic equtions hve three solutions, ll qurtic equtions hve four solutions, nd so on. For ll polynomil equtions with rel coefficients, ny complex roots lwys occur in conjugte pirs. So the three roots of cubic eqution could be three rel roots (possibly including repeted roots) or one rel root nd conjugte pir of complex roots. Similrly, the four roots of qurtic eqution could be four rel roots (possibly including repeted roots), two rel roots nd conjugte pir of complex roots, or two conjugte pirs of complex roots. However, in Further Pure 1 you will only be required to solve qudrtic equtions with complex roots. Equting rel nd imginry prts For two complex numbers to be equl, then the rel prts must be equl nd the imginry prts must be equl. So one eqution involving complex numbers cn be written s two equtions, one for the rel prts, one for the imginry prts. The exmples below show how this technique cn be used to solve equtions involving complex numbers. Exmple 5 Solve the eqution 3z z* = 4 15i Let z = x+ iy 3z z* = 4 15i 3( x+ i y) ( x i y) = 4 15i 3x+ 3iy x+ iy = 4 15i x+ 5iy = 4 15i Equting rel prts: x = 4 Equting imginry prts: 5y = 15 y = 3 z = 4 3i MEI, 17/03/06 5/7
In Exmple 5 the vlues of x nd y could be found immeditely by equting rel nd imginry prts. In the next exmple, equting rel nd imginry prts gives you pir of liner simultneous equtions to solve. Exmple 6 Solve the eqution (3 i)(z 1 + 4i) = 7 + 4i Let z = x + iy (3 i)(x + iy 1 + 4i) = 7 + 4i (3 i)((x 1) + i(y + 4)) = 7 + 4i 3(x 1) i(x 1) + 3i(y + 4) i²(y + 4) = 7 + 4i 3(x 1) i(x 1) + 3i(y + 4) + (y + 4) = 7 + 4i Equting rel prts: 3(x 1) + (y + 4) = 7 3x + y = Equting imginry prts: -(x 1) + 3(y + 4) = 4 -x + 3y = -10 6x + 4y = 4 3-6x + 9y = -30 Adding: 13y = -6 y = -, x = z = i Exmple 7 Find rel numbers nd b with > 0 such tht ( + bi)² = 16 30i. ( + bi)² = 16 30i ² + bi + b²i² = 16 30i ² + bi b² = 16 30i 15 Equting imginry prts: b = 30 b = Equting rel prts: ² b² = 16 5 Substituting: = 16 4 Multiplying through by ²: 5 = 16 4 16 5 = 0 This is qudrtic in ² nd cn be fctorised: (² 5)(² + 9) = 0 Since is rel, ² + 9 cnnot be equl to zero. Since > 0, the only possible solution is = 5. 15 15 b= b= = 3 5 = 5, b = 3 MEI, 17/03/06 6/7
Notice tht in this exmple you re finding the squre root of complex number. The squre root of 16 30i is 5 3i. You cn find the other squre root of this complex number by llowing to be negtive, so = -5 nd then b = 3. So the other squre root is -5 + 3i. As with rel numbers, one squre root is the negtive of the other. MEI, 17/03/06 7/7