An Introduction to Complex Numbers - A Complex Solution to a Simple Problem ( If i didn t exist, it would be necessary invent me.

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1 An Intrductin t Cmple Numbers - A Cmple Slutin t a Simple Prblem ( If i didn t eist, it wuld be necessary invent me. ) Our Prblem. The rules fr multiplying real numbers tell us that the prduct f tw negative numbers is psitive, fr instance ( 1)( 1) = 1. Thus the square f any real number, psitive r negative, is never negative. This fact can be summarised in symbls as fllws: If R, then 0. In particular, there is n real number such that = 1, r equivalently, such that + 1 = 0. This harmless-lking quadratic equatin has n slutins; at least, there are n real numbers which satisfy it. This was a prblem fr mathematicians right up t the 17th century, and it is nw ur prblem in these ntes. Histrical Reflectins An Ancient Prblem. The mathematicians f classical Greece, wh lived ver 000 years ag, had a similar difficulty with the s-called real numbers. They accepted fractins as numbers after all they culd represent them as lengths cnstructed with straight-edge and cmpass but they had truble with a number like which they knew was nt a fractin 1. They als knew frm Pythagras s Therem that was the length f the diagnal f a unit square (a square with sides f length 1), and s if numbers are t represent length, there has t be a number whse square is, a number which cannt be epressed as a fractin. (Numbers epressible as psitive r negative fractins are called ratinal numbers, and the set f all ratinal numbers is dented by the special symbl Q. Our prf in the ftnte belw shws that is irratinal.) An Imaginary Prblem. We nw knw hw t apprimate as clsely as we please because there are algrithms which generate its decimal epansin t as many places as we like (pcket calculatrs use such an algrithm). In fact, the real numbers were created in lgically satisfactry ways by Cauchy and Dedekind ver 100 years ag. We shall say a little mre abut their cnstructin later in the curse, and yu will learn t eplit their prperties. The prblem mathematicians had with the square rt f minus ne was a psychlgical ne; they had a cmple abut numbers being real, they wrried a lt abut whether they eisted. Nwadays we realise that even the s-called real numbers are a human inventin, a prduct f ur imaginatin, and are n mre real than angels dancing n the head f a pin. They make a very gd mdel fr handling length in gemetry, but they are nly a mdel, an abstract representatin. The sense f their being real is a trick f the brain and cmes frm identifying the numbers with the perceived reality f length. Of curse, 1 des nt represent a length, but then neither des 1 itself and yet n ne denies the usefulness f negative numbers this pint was made frcefully as lng ag as 1673 by the British mathematician Wallis. In fact, we shall see belw that the cmple numbers prvide a gd mdel fr -dimensinal space (i. e. the plane). Therefre let us ignre the questin f the eistence f 1 and base ur judgement n whether the cncept is useful. Utility is a mre imprtant prperty than reality (whatever we mean by that). A Slutin, hedged abut with dubts. Diphantus (c.75 AD) attempted t slve the plausible prblem f finding a right-angled triangle with perimeter 1 and area 7 and derived a quadratic equatin fr the length f a side which had n real rts (s n such triangle eists). The Italian mathematician Pacili stated in 1494 that the equatin + c = b cannt be slved 1 If culd be written as a fractin a/b, we culd als suppse that a and b are nt bth even (therwise we culd cancel a cmmn factr f frm a and b). Observe that the square f an dd number is always dd. If the square f a/b were, it wuld fllw that a = b, s a culd nt be dd. But then a, being even, is twice sme whle number c whence a = (c) = 4c, which means that 4c = b and therefre b = c. Nw by the same reasning b, like a, must be even, which we suppsed nt t be the case. This cntradictin means that is nt a fractin. (This is an eample f prf by cntradictin.) 1

2 unless b 4c. Anther Italian mathematician Girlam Cardan described the equatin = 6 as impssible, referring t the rts as fictitius. Hwever, he was willing t use square rts f negative numbers t divide 10 int tw parts whse prduct is 40 thus: 10 = (5 + 15) + (5 15). Karl Friedrich Gauss was the first t call epressins f this kind cmple numbers. Cmple numbers gradually established themselves as a valuable etensin t the real number system and althugh dubts abut their eistence slwly disappeared, the legacy f the ld terminlgy real and imaginary still survives. By the middle f the last century the thery f cmple variables was a thriving and central branch f mathematics. Nw let s g t it! The Answer t ur Prblem Definitin. A cmple number is an epressin f the frm + iy, where and y are real numbers and i is an algebraic symbl satisfying i = 1. The set f all cmple numbers is dented by the special symbl C. Remarks and Ntatin. 1. When y = 0 we write simply rather than + i0; in particular, 0 + i0 is dented by 0. The subset f C cnsisting f all + iy with y = 0 is a cpy f the set f real numbers sitting inside C. If we dente the set f real numbers by the special letter R, we can epress this symblically by R C. (Here the ntatin means is a subset f.) Frm this viewpint, a real number is a special case f a cmple number.. One f the structural rules f the cmple numbers is that + iy = 0 if and nly if = 0 and y = 0. (Aside: In the language f Linear Algebra, the cmple numbers 1 and i are linearly independent ver R.) 3. Tw mre structural rules in C are: iy = yi and ( y)i = (yi). Putting them tgether, we can write a cmple number like 5 + i( ) as 5 i instead. 4. It is ften cnvenient t use a single letter, such as z, t stand fr a typical cmple number + iy. Tw Remarkable Facts I. In C we can (i) add, (ii) subtract, (iii) multiply, and (iv) divide by nn-zer cmple numbers in such a way that the familiar rules f arithmetic are satisfied, fr eample the Distributive Law, which states that u(v + w) is the same number as uv + uw fr all chices f cmple numbers u, y, and w. (A set which admits these fur basic algebraic peratins is called a field. Thus Q (the ratinal numbers), R (the real numbers), and C are all eamples f fields.) II. Every plynmial equatin (nt just + 1 = 0) nw has a rt in C, even when the cefficients f the plynmial are themselves cmple numbers. This fact, tgether with the remainder therem, implies that a plynmial f degree n has eactly n rts in C (cunting repetitins). The secnd remarkable fact is t deep t prve at this stage, but the first is straightfrward, as we nw see. Prf f Remarkable Fact I Additin and subtractin f cmple numbers is easy. Let z = + iy and z = + iy, and define the fllwing peratins. Sum: z + z = ( + iy) + ( + iy ) = ( + ) + i(y + y ) Difference: z z = ( + iy) ( + iy ) = ( ) + i(y y ) Nte that in each case the right-hand side has the frm f a real number plus i times anther real number. Thus the sum and difference f tw cmple numbers is anther cmple number. We say that the set C is clsed under the peratins f additin and subtractin. Multiplicatin is nt difficult, prvided we remember t replace i by 1 whenever we can. We will dente the prduct f tw cmple numbers z and z by zz rather than by z z. Jutapsitin (writing tw symbls net t each ther) is ften used by mathematicians t dente sme kind f prduct r multiplicatin. The usual rules fr epanding brackets give:

3 Prduct: zz = ( + iy)( + iy ) = + iy + iy + iyiy = + i(y + y ) + i yy = ( yy ) + i(y + y ) Since,, y, and y are all real numbers, the tw numbers ( yy ) and (y + y ) are als real. Thus the prduct we have just defined (the final epressin in the abve equatins) is again a cmple number. This shws that C is clsed under multiplicatin. Finally, we want t divide by a nn-zer cmple number z = + iy. T say that z is nn-zer means that at least ne f and y is nn-zer. T divide by z is the same as multipying by its inverse 1/z, and since we already knw hw t multiply, it will be enugh t identify a cmple number w equal t 1/z. Nw if w = 1/z, then zw = 1. The secret f finding such a w, is t ntice that if z (= + iy) is multiplied by its s-called cmple cnjugate z = iy, the answer is the psitive real number + y (why is it psitive?). T see this, apply the prduct rule abve with = and y = y: z z = ( + iy)( iy) = ( y( y)) + i(y + ( y)) = ( + y ) + i0 = + y. If we nw replace z by ( w = 1 + y ) ( ) ( ) y z = + y i + y and carry thrugh the same calculatin, we btain zw = 1. This yields the desired inverse 1/z (als dented by z 1 ) f a nn-zer cmple number z = + iy: Inverse: z 1 = 1 ( ) ( ) y z = + y i + y. This cmpletes ur prf that in C we can add, subtract, multiply pairs f cmple numbers, and als divide by nn-zer cmple numbers, thus shwing that C is a field. We mentined earlier that, just as the real numbers can be used t represent pints n a line, the cmple numbers are a useful mdel fr studying pints in the plane. A Gemetrical View f the Cmple Numbers The underlying idea is simple. We identify the cmple number z = + iy with the pint in the plane whse crdinates are (, y). We then interpret gemetrically the fur arithmetical peratins defined abve. If P is the pint represented by the cmple number z = + iy, its crdinates are (, y), and the distance r f P frm the rigin O is + y by Pythagras s therem. If the Greek letter θ (theta) dentes the angle between the -ais and OP (measured anticlckwise), we have = r cs θ and y = r sin θ, and therefre z = r(cs θ + i sin θ), where r = + y. Therefre the pint with plar crdinates (r, θ) is represented by the cmple number r(cs θ + i sin θ). There are special names given t r and θ in this situatin. Definitins. (a) The mdulus f a cmple number z, dented by z, is defined thus: z = + y. (b) The argument f a cmple number z is the angle θ between the -ais and OP measured psitively in an anti-clckwise directin. By cnventin we chse the s-called principal value f the argument lying in the range 0 θ < π. Thus arg z = tan 1 (y/) with 0 arg z < π The sum f tw cmple numbers has a well-knwn gemetrical meaning. If P and P are pints represented by z = + iy and z = + iy, the crdinates f the pint represented by the sum 3

4 y imaginary ais r P z=+iy y O real ais Figure 1: THE ARGAND DIAGRAM z + z = ( + ) + i(y + y ) are ( +, y + y ), and these are the crdinates f the vectr sum OP + OP. Hence the sum f tw cmple numbers represents the vectr sum f their pints in the plane. Similarly the difference z z crrespnds t the difference f their crrespnding vectrs. The sum and difference f tw cmple numbers can therefre be visualised by means f the pictures shwn in figures and 3. The prduct f tw cmple numbers is best epressed in terms f their plar crdinates. Therem. Let z and z be cmple numbers with mduli r and r and arguments θ and θ respectively. In ther wrds, suppse that z = r(cs θ + i sin θ) and z = r (cs θ + i sin θ ). Then zz = rr (cs(θ + θ ) + i sin(θ + θ )). (1) Hence t calculate the prduct f tw cmple numbers, multiply their mduli and add their arguments. In symbls, this means: zz = z z and arg zz = arg z + arg z (minus π when necessary). Prf Using the prduct rule and the familiar frmulas fr sin(a + B) and cs(a + B), we get (cs θ + i sin θ)(cs θ + i sin θ ) = (cs θ cs θ sin θ sin θ ) + i(sin θ cs θ + cs θ sin θ ) = cs(θ + θ ) + i sin(θ + θ ) Equatin 1 nw fllws, and shws that the prduct zz can be epressed in the frm f a cmple number whse mdulus is rr = z z and whse argument is θ + θ, prvided we subtract π t btain the principle value when θ + θ falls in the range π t 4π. The rule fr the prduct stated in the abve Therem makes it very easy t describe the inverse. If we multiply the cmple number z crrespnding t (r, θ) by the cmple number z crrespnding t (r 1, θ), we btain the cmple number whse mdulus is r r 1 = 1 and whse argument is θ + ( θ) = 0, and this is just the number 1 + i0 = 1. In ther wrds, zz = 1 and hence z = z 1. We have therefre shwn that if z = r(cs θ + i sin θ), then z 1 = r 1 (cs θ i sin θ) 4

5 y imaginary ais z z+z z real ais Figure : The sum f tw cmple numbers y imaginary ais z z z z real ais z Figure 3: The difference f tw cmple numbers 5

6 y imaginary ais z= (cs(60)+isin(60)) real ais 1 z =0.5(cs(60) isin(60)) since sin( θ) = sin θ. The abve diagram shws the inverse z 1 f the fllwing cmple number: ( π ) ( π ) z = (cs + i sin = 1 + i de Mivre s Therem Using the multiplicatin rule described abve repeatedly fr cmple numbers we can shw that fr a cmple number z = r(cs θ + i sin θ), z n = r n (cs nθ + i sin nθ) This result is knwn as de Mivre s Therem. Fr eample z = r (cs(θ + θ) + i sin(θ + θ)) and z 3 = zz = r 3 (cs(θ + θ) + i sin(θ + θ)) = r 3 (cs 3θ + i sin 3θ) etc. de Mivre s Therem is valid fr any integer r ratinal n. The epnential frm f a cmple number In ur gemetrical view f cmple numbers we see that the cmbinatin cs θ + i sin θ invlving the angle θ keeps crpping up. It is cnvenient and very useful t find a mre cncise epressin fr this cmbinatin. Let s cnsider differentiating z = cs θ + i sin θ with respect t θ, dz dθ = sin θ + i cs θ and again d z dθ = cs θ i sin θ which is f curse z. Nw we will repeat this prcess with the cmple number z 1 = e iθ dz. 1 dθ = ieiθ and d z 1 dθ = i e iθ which is z 1. This suggests that cs θ + i sin θ e iθ (This result can be shwn frmally by using pwer series epansins. cs θ can be epressed as a series and similarly fr sin θ cs θ = 1 θ! + θ4 4! θ6 6! + sin θ = θ θ3 3! + θ5 5! θ7 7! + fr any real value f θ. (The factrial n! = 1..3 n, e.g. 4! = ) 6

7 The epnential functin e can be epressed fr any real as and if we define e z as e = 1 + +! + 3 3! + 4 4! + e z = 1 + z + z! + z3 3! + z4 4! + where z is a cmple number, it is easy t shw that e iθ = 1 + (iθ) + (iθ)! + (iθ)3 3! + (iθ)4 4! + which is equivalent t cs θ + i sin θ when i = 1 has been used t tidy up this epressin.) Thus a cncise way f writing any cmple number z = r(cs θ + i sin θ) is z = re iθ. The cmple cnjugate z = re iθ and if we revisit de Mivre s Therem The relatins and z n = (re iθ ) n = r n e inθ = r n (cs nθ + i sin θ) cs θ = eiθ + e iθ sin θ = eiθ e iθ i are als handy and ften used. Eample Find three different cmple numbers which satisfy the equatin z 3 = 1. Epress z as re iθ and find values f r and θ that satisfy the equatin. becmes Equating real and imaginary parts gives (re iθ ) 3 = 1 r 3 e 3iθ = 1 r 3 cs 3θ = 1 r 3 sin 3θ = 0 We find r = 1 frm squaring and adding the equatins and cs 3θ = 1 and sin 3θ = 0. This means that 3θ = 0, π, 4π, and θ = 0, π/3, 4π/3, π, π/3 + π, 4π/3 + π,. The three cmple numbers which satisfy z 3 = 1 are thus z 1 = 1 z = cs π 3 = i sin π 3 = 1 + i 3 z 3 = cs 4π 3 = i sin 4π 3 = 1 i 3 7

8 Eercises ATTEMPT THESE QUESTIONS AND HAND IN YOUR ANSWERS TO QUESTIONS 1 TO 8 TO YOUR PERSONAL TUTOR BY THE END OF THE FIRST WEEK OF TERM. 1. Write the fllwing cmple numbers in the frm + iy, where and y are real: (i) ( 1 i)(3 +4i); (ii) ( + i) ( i) (iii) (1 + i)/(1 i); (iv) (9 + 3i)/(6i) (1 mark). Fr what real value f a is the equatin (1 + 3i)(5 ai) = 50 satisfied? (1 mark) 3. On the Argand diagram mark the pints represented by the fllwing cmple numbers: (i) 4i; (ii) 1 i; (iii)(1 + i)3. (1 mark) 4. Find the mdulus and argument f each f the fllwing cmple numbers: (i) 3 i; (ii) 6 + 8i; (iii) i. (1 mark) 5. Mark the fllwing pints n the Argand diagram and verify the prduct rule described in the ntes: (a) 3 + i; (b) + i; (c) their prduct. (1 mark) 6. Shw that the fllwing fur pints frm the vertices f a square in the Argand diagram: 4 + 3i, 3 + 4i, 4 3i, 3 4i. (1 mark) 7. Wrk ut ( 1 + i. 1 ) 4, in ther wrds, epress this furth pwer in the standard frm + iy. Find fur different cmple numbers z satisfying z = 0. ( marks) 8 Find tw cmple numbers which satisfy the equatin (3+i) z = 1 and mark their psitins n the Argand diagram.( marks) (Optinal) Give a sensible meaning t the symbl i i, that is t say, i raised t the pwer f i. (Hint: Use Euler s beautiful identity cnnecting the famus five numbers 0, 1, e, π and i, namely e iπ + 1 = 0.) 8

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