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.)

3 2 +i=(2,1)for 2 +i=x+iyon (x,y)-plane)Two methods: (x,y) the pointPon the (x,y)-planeii. 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.

4 (d)3iBy inspection, 3i=3 cos 2+isin 2 . (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.

5 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. 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.

6 , 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).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.

7 |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).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).

8 |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. 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.

9 (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. 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.

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 . 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 ))