Complex Numbers Summary

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Bertina Gertrude Singleton
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1 Complex Numbers Summary What does a complex umber mea? Academic Skills Advice A complex umber has a real part ad a imagiary part (the imagiary part ivolves the square root of a egative umber). We use Z to deote a complex umber: e.g. Z = x + iy Imagiary Z = + 3i Re(Z) = Im(Z) = 3 You might see the i before or after it s umber - it does t matter which. Sometimes (especially i egieerig) a j is used istead of i they mea the same thig. Powers of i i stads for 1 so: i 2 = ( 1) 2 = -1 i = (i 2 ) 2 = (-1) 2 = 1 For ay power of i substitute as may i 2 s as possible ad they will all ed up as ±i or ±1. Examples: i 2 = (i 2 ) 1 = ( 1) 1 = 1 i 33 = (i 2 ) 16 i = ( 1) 16 i = i Summary: i = 1 i 2 = 1 i = 1 Addig & Subtractig This is easy just add or subtract the real part ad add or subtract the imagiary parts: Examples: ( + 3i) + (2 + 6i) = (6 + 9i) (3 + 7i) (1 3i) = (2 + 1i) Multiplyig Multiply out the 2 brackets (ad remember that i 2 = 1). (3 + 5i)( 2i) = 12 6i + 2i 1i 2 = i 1 ( 1) = i H Jackso 21 / 215 / 216 / Academic Skills 1 Except where otherwise oted, this work is licesed uder

2 Complex Cojugate The cojugate is exactly the same as the complex umber but with the opposite sig i the middle. Whe multiplied together they always produce a real umber because the middle terms disappear (like the differece of 2 squares with quadratics). ( + 6i)( 6i) = 16 2i + 2i 36i 2 Complex umber Complex cojugate = 16 36( 1) = = 52 Dividig Dividig by a real umber: Dividig by a complex umber: divide the real part ad divide the imagiary part. Multiply top ad bottom of the fractio by the complex cojugate of the deomiator so that it becomes real, the cotiue as above. Examples: 3+i 2 = 3 + i = i 2 2 5i = 5i 3 2i = 12 8i 15i+1i2 = 12 23i+1( 1) = 2 23i = 2 23 i 3+2i 3+2i 3 2i 9 6i+6i i 2 9 ( 1) Graphical Represetatio A complex umber ca be represeted o a Argad diagram by plottig the real part o the x-axis ad the imagiary part o the y-axis. Imagiary x P (z = x + yi) y NB ta() = y x = ta 1 ( y x ) so * Whe fidig either: do a sketch to see where the agle is or remember: if x < (i.e. egative) add 18 (or π) to your aswer. Modulus: is writte as z or r ad is the legth of OP, therefore z = r = x 2 + y 2 Argumet: is the agle that is made with the horizotal axis (deoted by ). P (z = + 5i) 5 z = = 1 = 6. = ta 1 ( 5 ) =. 9 radias H Jackso 21 / 215 / 216 / Academic Skills 2

3 Polar & Expoetial Form As well as the cartesia form (z = x + iy) there are 2 alterative ways of writig a complex umber: Polar: Expoetial: z = re i z = r(cos + isi) Where: r is the modulus (legth of the lie) z is the argumet (agle it makes with the x-axis) (this should be i radias for the expoetial form). Remember: To fid the modulus (legth), r: use Pythagoras To fid the argumet (agle), : use ta 1 ( y x ) * Covertig betwee the differet forms: Cartesia Polar or Expoetial Need to fid r ad r = x 2 + y 2 = ta 1 ( y x ) * Polar or Expoetial Cartesia Need to fid x ad y x = rcos y = rsi imagiary Express z = 3 + i i polar ad expoetial form Modulus: r = = = 25 = 5 r Argumet: = ta 1 ( 3 ) = 53.1o (i radias =.927 r ) 3 real Polar form: Exp form: z = 5(cos(53.1) + isi(53.1)) z = 5e.927i NB It is advisable to do a quick sketch of the complex umber ad check that the agle you calculated matches the diagram. If it s i a differet quadrat adjust the agle as ecessary. Remember that agles are positive whe measured aticlockwise ( ) Express z = 7e iπ 3 i cartesia form x = rcos x = 7 cos ( π 3 ) = 3.5 y = rsi y = 7 si ( π 3 ) = 6.1 Cartesia form: z = i H Jackso 21 / 215 / 216 / Academic Skills 3

4 A remider of the 3 forms: Cartesia Polar Expoetial z = x + iy z = r(cos + isi) z = re i Coversios: x = rcos r = x 2 + y 2 y = rsi = ta 1 ( y x ) * Multiplyig with Polar or Expoetial form Let z 1 = z 2 z 3 The z 1 = z 2 z 3 Ad z 1 = z 2 + z 3 This meas, whe multiplyig 2 complex umbers: Multiply the r s Add the agles ( s) If z 1 = 5e π 2 i ad z 2 = 3e π 3 i fid z 1 z 2 New modulus: 5 3 = 15 New agle: π 2 + π 3 = 5π 6 z 1 z 2 = 15e 5π 6 i Dividig with Polar or Expoetial form Let z 1 = z 2 z 3 The z 1 = z 2 z 3 Ad z 1 = z 2 z 3 This meas, whe dividig 2 complex umbers: Divide the r s Subtract the agles ( s) if z 1 = 5 (cos ( π 2 ) + isi (π 2 )) ad z 2 = 3 (cos ( π 3 ) + isi (π 3 )) fid z 1 z 2 New modulus: 5 3 = 5 3 New agle: π 2 π 3 = π 6 z 1 z 2 = 5 3 (cos (π 6 ) + isi (π 6 )) H Jackso 21 / 215 / 216 / Academic Skills

5 De Moivres Theorem: Is used for raisig a complex umber to a power. z = r (cos() + isi()) Thik: Raise r to the power of ad multiply the agle by. e.g If z = 3 (cos ( π 3 ) + isi (π 3 )) the z 5 = 3 5 (cos 5π 3 + isi 5π 3 ) If the power is a fractio (root) there are solutios: z 1 = r 1 (cos ( +2kπ Just has a 2kπ extra ) + isi ( +2kπ )) Remember that roots ca be writte as fractios: z = z 1 Where k =,1,2.-1 If you struggle to remember the theorem for whe the power is a fractio just use the origial (the 1 st equatio) exactly the same but remember that there will be solutios, all with the same modulus but with differet argumets. I this case, to fid the argumets you eed to keep addig 2π e.g. Fid the fifth roots of z = 2(cos(π) + isi(π)) Usig the origial equatio: to your previous aswer. Raise r to the power of ad multiply the agle by. 1 st aswer: z 1 5 = (cos ( π 5 ) + isi (π 5 )) (Remember there are 5 aswers so keep addig 2π 5 to fid the other ) 2 d aswer: z 1 5 = (cos ( 3π 5 ) + isi (3π 5 )) 3 rd aswer: z 1 5 = (cos ( 5π 5 ) + isi (5π 5 )) th aswer: z 1 5 = (cos ( 7π 5 ) + isi (7π 5 )) 5 th aswer: z 1 5 = (cos ( 9π 5 ) + isi (9π 5 )) H Jackso 21 / 215 / 216 / Academic Skills 5

Complex Numbers. Brief Notes. z = a + bi

Complex Numbers. Brief Notes. z = a + bi Defiitios Complex Numbers Brief Notes A complex umber z is a expressio of the form: z = a + bi where a ad b are real umbers ad i is thought of as 1. We call a the real part of z, writte Re(z), ad b the