The usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1

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1 Mth50 Introduction to Differentil Equtions Brief Review of Complex Numbers Complex Numbers No rel number stisfies the eqution x =, since the squre of ny rel number hs to be non-negtive. By introducing new imginry number i =, which is supposed to hve the property i =, the rel number system cn be extended to lrge number system, nmely, the complex number system. A complex number is n expression of the form x + iy, where x, y R re rel numbers. x is clled the rel prt of the complex number, nd y the imginry prt, of the complex number x + iy. The rel prt nd imginry prt of complex number re sometimes denoted respectively by Re(z) = x nd Im(z) = y. Two complex numbers re equl, + ib = c + id, where, b, c, d R re ll rel numbers, if nd only if = c nd b = d. The usul lgebric opertions +,, (or ), on rel numbers cn then be extended to opertions on complex numbers in nturl wy: ddition ( + ib) + (c + id) = ( + c) + i(b + d) subtrction ( + ib) (c + id) = ( c) + i(b d) Exmple. multipliction ( + ib)(c + id) = (c bd) + i(d + bc) + ib c + bd + bc division = c + id c + i d + d c + d (3 + 5i) + ( + i) = (3 ) + (5 + )i = + 6i (3 + 5i)( + i) = (3 ( ) 5 ) + (3 + 5 ( ))i = 7i 3 + 5i + i = 3 ( ) ( ) ( ) + + ( ) + i = i For multipliction nd division of complex numbers, insted of using their forml definitions, it is more convenient to simply crrying out the lgebric opertions by following the usul lgebric rules, such s commuttive lw, ssocitive lw nd distributive lw. Exmple. i =, i 3 = i i = ( )i = i, i 4 = i 3 i = i i =, i 5 = i 4 i = i,. i 4n =, i 4n+ = i, i 4n+ =, i 4n+3 = i, where n = 0,,, 3, Exmple.3 (Multipliction) (3 + 5i)( + i) = (3 + 5i) ( ) + (3 + 5i) i = 3 ( ) + 5i ( ) + 3 i + 5i i Exmple.4 (Division) = 6 0i + 3i 5 = 7i The fct tht ( + ib)( ib) = + b leds to usul trick to find the quotient of two complex numbers: 3 + 5i (3 + 5i)( i) 6 3i 0i 5i = = + i ( + i)( i) ( ) i = i

2 q Conjugte, Modulus nd Polr Representtion of Complex Numbers If z = x + iy, where x, y re rel numbers, then its complex conjugte z is defined s the complex number z = x iy. It is esy to check tht (z + z) = x = Re(z) nd (z z) = iy = iim(z). Moreover, z z must be nonnegtive rel number, since z z = (x + iy)(x iy) = x + y The modulus or bsolute vlue of complex number z = x + iy is defined nd denoted by z = x + y = z z Exmple. Let z = 3i. Then z = + 3i nd z z = + 3 = 3. Hence + 3i = 3. Exercise Show tht () z z = z z, (b) Exmple. (Qudrtic Equtions with Rel Coefficients) z z = z z. The roots of qudrtic eqution z +bz+c = (z+ b ) +c b 4 = 0, where, b, c re rel numbers, re given by z = b ± b 4c two distinct rel roots if b 4c > 0 one repeted rel roots if b 4c = 0 two distinct (conjugte) complex roots if b 4c < 0 For exmple, z + z + = 0 hs two complex roots + 3 i to ech other. nd 3i, which re conjugte A complex number z = x + iy cn be grphiclly represented by point with coordintes (x, y) in the Crtesin coordinte plne. y r x + i y (x,y) x Using polr coordintes of the plne, i.e., letting x = r cosθ, y = r sin θ complex number z = x + iy 0 cn be expressed in the form z = r(cos θ + i sin θ)

3 where r = x + y = z is the modulus, nd θ is n rgument of z. Since sin nd cos hve period π, the rgument of z is defined up to n integrl multiple of π: rgz = θ + nπ, n = 0, ±, ±, Exmple.3 (Multipliction nd Division vi Polr Representtions) Note tht for z = r (cos θ + i sin θ ) nd z = r (cos θ + i sin θ ), we hve z z = r r (cosθ + i sinθ )(cos θ + i sinθ ) = r r [(cosθ cosθ sinθ sinθ ) + i(cosθ sin θ + cosθ sinθ )] = r r [(cos(θ + θ ) + i sin(θ + θ )] z = r (cosθ + i sinθ ) z r (cosθ + i sinθ ) = r (cos(θ θ ) + i sin(θ θ )] r In prticulr, De Moivre s Theorem follows: z n = r n (cosnθ + i sin nθ) for ny integer n. For exmple, (cosθ + i sinθ) = cosθ + i sin θ, nd (cosθ + i sin θ) = cos( θ) + i sin( θ) = cosθ i sinθ. 3 Complex Exponentil Function The rel-vlued exponentil function e x cn be extended to complex function e z by the following definition: e z = e x+iy = e x e iy = e x (cosy + i sin y) Using the complex exponentil function, the polr representtion of complex number cn be expressed s z = r(cos θ + i sinθ) = re iθ, where θ is choice of the rgument of z. It is esy to see from the definition tht e z+πi = e x+i(y+π) = e x (cos(π + y)+isin(y + π)) = e z, i.e., e z is complex-vlued function with complex period πi. Exmple 3. e iπ = cosπ + i sinπ = ; e 3+ π 4 i = e 3 e π 4 i = e 3 (cos π 4 + i sin π 4 ) = e 3 + e 3 i. Exmple 3. It is not hrd to show from the definition: e z e z = e z+z. Exmple 3.3 The roots of the eqution z n = for ny positive integer n cn be found from the polr representtions z n = = re iθ = re i(kπ+θ), k = 0,,,... s z = r n e kπ+θ n i, k = 0,,,..., n. For exmple, the roots of z 3 = = e kπi re:, e πi 3, e 4πi 3, i.e.,, + 3 i, 3 i. Exmple 3.4 (Complex Exponentil Functions of Rel Vrible) For ny rel number, b, nd rel vrible t, the complex exponentil functions e (+ib)t nd e ( ib)t of the rel vrible t simply men: e (+ib)t = e t (cosbt + i sinbt), e ( ib)t = e t (cos bt i sinbt). 3

4 Note tht in prticulr, e t cosbt = Re(e (+ib)t ) = e(+ib)t + e( ib)t e t sin bt = Im(e (+ib)t ) = i e(+ib)t i e( ib)t For exmple, e (+3i)t is the function e t cos3t + ie t sin3t. Similrly, e t cost nd e t sin t re the rel nd imginry prt of the complex-vlued exponentil function e ( +i)t respectively. 4 Complex-vlued Functions of Rel Vrible If f(t) nd g(t) re functions of rel vrible t, complex-vlued function h(t) of rel vrible cn be defined by tking h(t) = f(t) + ig(t). Then f(t) = Re h(t), g(t) = Im h(t) The derivtive or integrls of h(t) cn be defined in strightforwrd wy by h(t)dt = f(t)dt + i h (t) = f (t) + ig (t) g(t)dt, b h(t)dt = b f(t)dt + i A useful complex-vlued function for solving liner differentil eqution is Direct computtion shows h(t) = e (+ib)t = e t (cos bt + i sinbt) b g(t)dt de (+ib)t dt = ( cosbt b sin bt)e t + i( sinbt + b cosbt)e t = ( + ib)e (+ib)t i.e., the ordinry derivtive formul for e t extends to the complex cse e (+ib)t. Exmple 4. (I) de (3 i)t dt = (3 i)e (3 i)t, nd hence e (3 i)t dt = 3 i e(3 i)t + c = 3 + i 3 e3t (cost i sin t) + c, where c is n rbitrry complex integrtion constnt. Now, since (II) e (3 i)t dt = e 3t cos( )tdt + i e 3t sin( )tdt = e 3t costdt i e 3t sin tdt, by comprison of rel nd imginry prts on the right sides of (I) nd (II), one gets bck the rel integrls: e 3t costdt = e 3t( 3 3 cost + 3 sin t) + C, e 3t sin tdt = e 3t( 3 sin t 3 3 cost) + C 4

5 Exercise. Express the following complex numbers in the form of x + iy: () ( 5i) + (3 + 4i) (b) ( + 3i) (3 + i) (c) (3 5i)( + i) (d) ( 3i) 3 (e) 3 + i (f) 5 3 5i 4 + 3i. Find find z, z nd s lest one polr representtion of z for the following complex numbers: () z = i (b) z = 3i (c) z = + 3i 3i 3. Find ll the roots of the qudrtic eqution z 3z + 3 = Express the following complex numbers in the form of x + iy: () e πi 4 (b) e 3πi 4 (c) e 3πi 5. Find dy dt nd y(t)dt for the following functions: () y(t) = e (3+i)t (b) y(t) = e it (c) y(t) = e ( +i)t 6. Use suitble complex exponentil function to integrte e t cos3tdt. 5