Multiplication:z1×z2=r1×r2cisϑ1+ϑ2Division:z1z2=r1r2cisϑ1-ϑ2Example.Euler’s and De Moivre’s theoremThese two theorems state the relationship between the trigonometric functions and thecomplex exponential function. This allows us to convert between Cartesian and Polarforms.Euler’s Theoremeix=cosx+isinxDe Moivre’s theoremzn=r(cosx+isinx)n=rncos(nx) +isin(nx)De Moivre’s theorem can be derived from Euler’s through theexponential law for integer powers.(eix)n=eixn=zn20
ALGEBRAComplex numbers1De Moivre’s theorem: proof by inductionHaving seen the method of induction, we will now apply it to De Moivre’s theorem.Prove:zn=cos(x)isin(x)n=cos(nx) +isin(nx).1. Show true forn=1cos(x)isin(x)1=cos(1x) +isin(1x)cos(x)isin(x) =cos(x) +isin(1x)is true forn=12. Assume true forn=kcos(x) +isin(x)k=cos(kx) +isin(kx)3. Prove true forn=k+1cos(x) +isin(x)k+1=cos(k+1)x+isin(k+1)x=cos(x) +isin(x)1cos(x) +isin(x)kInductive step: use assumption aboutn=k=cos(x) +isin(x)cos(kx) +isin(kx)Remember i2=-1=cos(x)cos(kx) +cos(x)isin(kx) +isin(x)cos(kx)-sin(x)sin(kx)=cos(x)cos(kx)-sin(x)sin(kx) +i cos(x)sin(kx) +sin(x)cos(kx)Use of the double/half angle formulae=cosϑ+kϑ+isinϑ+kϑ=cos(k+1)ϑ+isin(k+1)ϑis required result and form.Hence, by assumingn=ktrue,n=k+1 is true. Since the statement is true forn=1, it is true for alln∈Z+.Example.21
ALGEBRAComplex numbers1.5.3Nth roots of a complex numbernthroot of a complex numberzis a numberωsuch thatωn=z.To findnthroot of a complex number (in polar form,z=rcis(ϑ)) you need to use thefollowing formula:zk+1=npr×cisϑn+2kπnfork=0,1,2,...,n-1.Finding complex rootsFind 3 roots ofz3=4+4p3iand draw them on the complex plane.1.Rewrite the complex number inpolar form.r=¨42+4p32=p64=8ϑ=arctanp3=π3z3=8cisπ32.Insert values into formula.z1=3p8×cisπ3×3=2cisπ9z2=3p8×cisπ3×3+2×1×π3=2cis7π9z3=3p8×cisπ3×3+2×2×π3=2cis13π93.Draw the roots on the complexplane, they should be equallyspaced out with the same length.z1z2z322
ALGEBRASystem of linear equations: Unique, infinite and no solutions11.6System of linear equations: Unique,infinite and no solutionsSolving a system of two linear equations should be familiar to most of you. There areseveral methods of solving it, including substitution and subtraction of equations fromeach other. However, sometimes there can be three equations with three unknowns oreven two equations with three unknowns. It is important to identify when thoseequations have unique, infinite amount or no solutions at all. The easiest way to do it isto solve the simultaneous equations. It is possible to think of a system of linear equationsgeometrically, where the solution is at the intersection of lines or planes. Thus theintersection can be a point, a line or a plane. Here is how to identify the amount ofsolutions that the system of equations has:Unique solutionthere is only one set of variables that satisfy all equations. Intersectionis a point.