Lecture 1 Complex Numbers - 4unitmaths.com

1 1. i= Numbers are often denoted asRis the set of real Numbers ,Cis the set of Complex Numbers .Ifzis a complexnumber,zis of the formz=x+iy C,for somex,y 3 + 4iis a Complex +iy real partimaginary +iy,x,y R,therealpartofz= (z) = Re(z)=xtheimaginarypartofz= (z) = Im(z)= +4i (z)=3 (z)= +iy,thenz( zbar ) is given byz=x iyand is called .Ifz=3+4i, thenz=3 2x+3= ( 2) ( 2)2 4(1)(3)2(1)=2 82=2 2 22=1 2i. Lecture 2 Complex 1.(2+3i)+(4+i)=6+ 2.(8 3i) ( 2+4i)=10 1.(2+3i)(1+2i)=2+4i+3i 6= 4+7iExample 2.(3 2i)(3+2i)=9 (2i)2=9+4=13 when we multiply two Complex conjugates, we get a real +3i1+4i=2+3i1+4i 1 4i1 4i=(2+3i)(1 4i)(1+4i)(1 4i)=2 8i+3i 12i21 (4i)2=14 5i17(realisingthe denominator) Lecture , ,ifa+ib=c+idwherea,b,c,d R,thena=candb= ,yif (3 + 4i)2 2(x iy)=x+ hand side (LHS) = 9 16 + 24i 2x+i2y= 7 2x+i(24 + 2y) 7 2x=x3x= 7x= 73&24+2y=yy= 24 Example ,yifx1+i+y2 i=2+ S=x1+i+y2 i=x1+i 1 i1 i+y2 i 2+i2+i=x(1 i)1+1+y(2 +i)4+1=x(1 i)2+y(2 +i)5 Nowx(1 i)2+y(2 +i)5=2+4i.

2 5x(1 i)+2y(2 +i) = 20 + 40i5x i5x+4y+i2y=20+40i5x+4y+i( 5x+2y) = 20 + 40iEquating real and imaginary part,5x+4y=20 5x+2y=40 Solving simultaneously,6y=60y=10& x= 4. Lecture 4 Square Roots of Complex the square root of 35 35 12i=a+ib: square both 12i=(a+ib)2=a2 b2+i(2ab) a2 b2=35and 2ab= 12ab= inspection, solutions area=6&b= 1ora= 6orb= b2=35ab= 6b= 6a. a2 6a 2=35a2 36a2= 36=35a2a4 35a2 36=0.(a2 36)(a2+1)=0a2=36 &a2+1=0 a/ R a= 6& b= 35 12i=6 i. (By convention,sign( ( z)) = sign( (z)))Example the roots ofz2 (1 i)z+7i 4 = 0 in the forma+ (1 i) (1 i)2 4(1)(7i 4)2=(1 i) 1 1 2i 28i+162=(1 i) 16 30i2 From beside,=(1 i) (5 3i)2=1 i+5 3i2or1 i (5 3i)2=3 2ior 2+i. 16 30i=(a+ib)16 30i=a2 b2+i(2ab)a2 b2=162ab= 30ab= 15a=5&b= 3ora= 5&b=3& 16 30i=5 3i sign(16) = sign(5) = + Lecture 5 The Argand Diagram.(Note: Ordered pairs:- eg. 2 +i=(2,1)for 2 +i=x+iyon (x,y)-plane)Two methods: (x,y) the pointPon the (x,y)-planeii.

3 Vector OPx-axis is called is called Plot the following on the Argand diagram:P=2+3i;B=3 i;A= 2 i;M=4;E=2iz=x+iy=rcos +irsin =r(cos +isin )Modulus(DistanceOP)denoted byr,modz,|z|,|x+iy|by Pythagoras,r2=x2+y2r= x2+y2r=|z|=|x+iy|= x2+ (angle )denoted by ,argz,arg(x+iy) [or ampz, amp (x+iy){amplitude}]by definition, 180 < 180 Forx =0,tan = mod-arg form of a Complex numberz=x+iy=r(cos +isin )(=rcis ). Complex ConjugateIfz=x+iy, then the Complex conjugate isz=x iyRadian measure(orcircularmeasure) =2 radians = 2 rad = 2 c=2 180 = 90 = 260 = 345 = 430 = 6 More on mod-arg the following in mod-arg form:-(a)2+2i;(b)2+5i;(c) 1+ 3i;(d)3i;(e)1 3i(a)2+2ir= 22+22= 8=2 2& tan =22=1& = 4& 2+2i=2 2 cos 4+isin 4 . (b)2+5ir= 22+52= 29& tan =52& = tan 152 68 12 & 2+5i= 29 cos tan 152 +isin tan 152 29 cos68 12 +isin68 12 (c) 1+ 3ir= 12+3= 4=2tan = 31& = 3& = 3=2 3& 1+ 3i=2 cos2 3+isin2 3 . (d)3iBy inspection, 3i=3 cos 2+isin 2.

4 (e)1 3ir= 12+32= 10tan( )=3& = tan 13& = tan 13 71 34 .& 1 3i= 10 cos tan 13 +isin tan 13 = 10 cos tan 13 isin tan 13 10 cos71 34 isin71 34 .Lecture6 AxiomsAnintegraldomainis a set of elements with two binary operations defined for them, whichobey the laws obeyed by setSis an integral domain if its elementsa,b,c,..obey the following Closure Law for Addition, ,a+b S2. Closure Law for Multiplication, ,a b S3. Commutative Law for Addition, ,a+b=b+a4. Commutative Law for Multiplication, ,a b=b a5. Associative Law for Addition, ,a+(b+c)=(a+b)+c6. Associative Law for Multiplication, ,a (b c)=(a b) c7. Distributive Law of Multiplication over Addition, ,a (b+c)=a b+a c8. There exists an additive identity (or zero element) 0, such that for everya,a+0=0+a=a(Note 0 S)9. There exists a multiplicative identity (or unity element) 1, such that for everya,a 1=1 a=a(Note 1 S)10. There exists an additive inverse (or opposite), a, for each memberaof the set suchthata+( a)=( a)+a= Cancellation Law.

5 Ifab=acanda = 0, thenb= , the set of the integers, is an integral elements of afieldFobey the above axioms 1-10 for integral domains, (wherea,b,care elements ofF) and instead of the cancellation law, there is a law about the existenceof a multiplicative inverse (or reciprocal):11 .Ifa 1and 1 are elements ofF, anda a 1=a 1 a= 1, wherea = 0, thena , the set of Complex Numbers is a additive inverse ofz=2+3iis z= 2 3iExample multiplicative inverse ofz=2+3iisz 1=12+3i=12+3i 2 3i2 3i =2 7( ) cos(A+B) = cosAcosB sinAsinBsin(A+B) = sinAcosB+ sinBcosAMod-arg (cos 1+isin 1)&z2=r2(cos 2+isin 2)then ifz1=z2thenr1=r2& 1= |z1z2|=|z1||z2|and arg(z1z2) = argz1+ argz2 2 . , for example:arg(z1z2) = 100 + 140 360 = 120 arg z1z2 = argz1 argz2 2 . (cos 1+isin 1)andz2=r2(cos 2+isin 2)thenz1z2=r1(cos 1+isin 1) r2(cos 2+isin 2)=r1r2(cos 1cos 2 sin 1sin 2+isin 2cos 1+isin 1cos 2)=r1r2(cos( 1+ 2)+isin( 1+ 2)) (see ( ) above)& |z1z2|=r1r2=|z1||z2|and arg(z1z2)= 1+ 2= arg(z1) + arg(z2).

6 Extended:arg(z1z2 zn) = argz1+ argz2+ + argzn 2 n.|zn|=|z|n(eg.,|z3|=|zzz|=|z||z||z|=|z| 3).and arg(zn)=nargz 2 k. 1zn =1|z|nand arg 1zn = arg 1 arg(zn)=0 nargz 2 k= nargz 2 the modulus and argument ofz=(2 i)(1 3i).|z|=|2 i||1 3i|= 22+12 12+32= 5 10= 50=5 (z) = arg(2 i) + arg(1 3i)= tan 112 tan 13 98 8 .Example ( 1+2i)(1 +i) 2 3i|z|=| 1+2i||1+i|| 2 3i|= 5 2 13= 10 13= (z) = arg( 1+2i) + arg(1 +i) arg( 2 3i) 285 15 360 = 74 45 Lecture 8 Triangle +iandz2= 1+2i, z1+z2=1+3i. Polygon of Complex z1=z2+( z1):Triangle Inequalities.|z1+z2| |z1|+|z2|:|z1 z2| |z1| |z2|:Example the triangle inequalities ifz1=2 3i,z2= 1+4i,z1+z2=1+i,z1 z2=3 7i.|z1|= 13|z2|= 17|z1+z2|= 2|z1 z2|= 58.|z1+z2| |z1|+|z2| 2 13 + 17 |z1 z2| |z1| |z2| 58 13 17 . triangle inequalities hold. Product of Complex triangleOQRis constructed similar to the point (1,0).Multiplication byi, 1, byi, rotation 90 (anticlockwise).

7 Multiplication by 1, rotation 180 by i, rotation 270 anticlockwiseLecture9 Geometric Representation of Locus forms:-|z z1|=arepresentsacircle, 1.|z|= 2.|z 3|= 3.|z i|= 4.|z 1 2i|=2|z (1+2i)|=2centre(1,2), 5.|z| 3(note:-iflessthan,itisinside,ifitisgrea terthan,itisoutside.)Example <|z| 7.|z| 4and0 argz (z) 2ifz=x+iy,then (z)=y(& 1 y 2)Example 9. 6<argz (z) 2and (z) 1 Example (z) 2or (z) 1 Example 12.|z| 4or0 argz 3 Lecture10 Using Algebra to Represent Locus ProblemsExample algebraically that|z 2 i|= 4 represents a circle with radius 4 unitsand centre (2,1).|z 2 i|=4. |x+iy 2 i|=4. |(x 2) +i(y 1)|=4. (x 2)2+(y 1)2=4. (x 2)2+(y 1)2= is a circle centre (2,1), radius 4 units. Example the curve:(i) (z2)=3(ii) (z2)=4.(i) (z2)=3 ((x+iy)2)=3 (x2 y2+2ixy)=3x2 y2=3.(ii) (z2)=4. 2xy=4. xy= in geometric terms, the curve described by 2|z|=z+z+ |z|=z+z+4. 2|x+iy|=x+iy+x iy+4.

8 2 x2+y2=2x+4=2(x+2). x2+y2=x+2. x2+y2=(x+2)2. x2+y2=x2+4x+4. y2=4x+4. sideways parabola at vertex ( 1,0).Example the locus of (z+iz)<2. (x+iy+i(x+iy))<2. (x+iy+ix y)<2. x y< +i&z2=2+3ifind the locus ofzif|z z1|=|z z2|.|x+iy (1 +i)|=|x+iy (2+3i)|. |(x 1) +i(y 1)|=|(x 2) +i(y 3)|. (x 1)2+(y 1)2= (x 2)2+(y 3)2.(x 1)2+(y 1)2=(x 2)2+(y 3) 2x+1+y2 2y+1=x2 4x+4+y2 6y+9. 2x+4y= |z z1|=|z z2|will always be a straight line. It will always be the perpendicularbisector of the interval ( ) (A+B) = sinAcosB+sinBcosA& cos(A+B) = cosAcosB Moivres Theorem.(cos +isin )n= cosn +isinn .Proof. (By mathematical induction forn=0,1,2,..)Step 1. Testn= (cos +isin )0= cos0 +isin0=1= it is true forn= 2. Assume true forn= , (cos +isin )k= cosk +isink .Test forn=k+ , (cos +isin )k+1& cos(k+1) +isin(k+1) = (cos +isin )k(cos +isin )1= (cosk +isink )(cos +isin )(since we have assumed it true forn=k)= cosk cos +isin cosk +isink cos sink sin = cosk cos sink sin +i(sin cosk + sink cos )= cos(k + )+isin(k + )(see ( )above)= cos(k+1) +isin(k+1) = 3.

9 If the result is true forn= 0, then true forn=0+1, ,n= 1. If the result istrue forn= 1, then true forn= 1 + 1, ,n= 2 ans so on for all nonnegative integersn Example :(a)(cos isin ) 4(b)(sin icos )7(c)(cos2 +isin2 )3(cos isin )4.(a)(cos isin ) 4= cos( 4 ) isin( 4 )= cos4 +isin4 (b)(sin icos )7=( icos + sin )7= i7(cos isin )7=i(cos7 isin7 )= sin7 +icos7 (c)(cos2 +isin2 )3(cos isin )4=(cos +isin )6(cos isin )4=(cos +isin )6(cos( )+isin( ))4=(cos +isin )6(cos +isin ) 4= (cos +isin )10= cos10 +isin10 Example in the formx+iy:(a) cos 2+isin 2 6(b) 1+ 3 10.(a)(cos 2+isin 2)6= cos6 2+isin6 2= cos3 +isin3 = 1+0i= 1 (b)(1 + 3)10= (2(cos 3+isin 3)10=210(cos10 3+isin10 3)=210 12 i 32 = 512 512i 3 Lecture12De Moivre s Theorem and the Argand 3+irepresent the following on the Argand Diagram:z,iz,1z, z,2z,z,z2+z,z3 zz= 2(cos 6+isin 6)z 1= (2(cos 6+isin 6)) 1=12(cos 6+isin 6)=12(cos 6 isin 6)2z= 4(cos 6+isin 6)z2= (2(cos 6+isin 6))2= 4(cos 3+isin 3)z3= (2(cos 6+isin 6))3= 8(cos 2+isin 2)Solution on next Identities and DeMoivre s cos 6 in terms of cos.)

10 Hence show thatx= cos(2k+1) 12wherek=0,1,2,3,4,5 is a solution to the equation 32x6 48x4+18x2 1 = 0 and hence deducethat cos 12= 6 +isin 6 = (cos +isin ) using Pascal s example,(a+b)2=a2+2ab+b2(a+b)3=a3+3a2b+3 ab2+b3(a+b)4=a4+4a3b+6a2b2+4ab3+b4(a+b)5 =a5+5a4b+10a3b2+10a2b3+5ab4+b5(a+b)6=a6+ 6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6cos 6 +isin = (cos +isin )6= cos6 + 6 cos5 isin + 15 cos4 (isin )2+ 20 cos3 (isin )3+ 15 cos2 (isin )4+6 cos (isin )5+(isin )6- from Pascal s Triangle= cos6 +6icos5 sin 15 cos4 sin2 20icos3 sin3 + 15 cos2 sin4 +6icos sin5 sin6 cos 6 = cos6 15 cos4 sin2 + 15 cos2 sin4 sin6 equating parts= cos6 15 cos4 (1 cos2 ) + 15 cos2 (1 cos2 )2 (1 cos2 )3= cos6 15 cos4 + 15 cos6 + 15 cos2 (1 2 cos2 + cos4 ) (1 3 cos2 + 3 cos4 cos6 )= cos6 15 cos4 + 15 cos6 + 15 cos2 30 cos4 + 15 cos6 1+3cos2 3 cos4 + cos6 = 32 cos6 48 cos4 + 18 cos2 1If cos 6 = 0, then 6 = 2, 3 2, 5 2, 7 2, 9 2, etc.