Complex Numbers : Solutions

1 Complex Numbers : SolutionsDavid SwensonExercise cartesian point is equivalent to the Complex number 6i?What about 2?Since 6i= 0+6i, we identifya= 0 andb= 6 ina+bi. Therefore, (a, b) becomes(0,6). Similarly, 2 gives us 2 + 0i, and the point( 2,0).Exercise the Complex numberz=a+ib, what is|z|in terms ofaandb? [Hint: think back to trigonometry.]If we look at figure 1, we see that|z|, which is denotedrin the figure, is thehypotenuse of a right triangle with side lengthsaandb. So, by the Pythagoreantheorem, we have|z|=r= a2+ +ib, what is arg(z) in terms ofaandb? For the specialcase of a real number (b= 0) what is arg(z)?This is again just a little trigonometry. We know that arg(z) is the angleidentified in figure 1, and we again use the fact that we have a right the length of the side opposite the angle, andathe side adjacent, wehave tan(arg(z)) =b/a, orarg(z) = tan 1(b/a).There is a subtlety to the special case of a real number . The obvious answeris that arg(z) = tan 1(0) = 0.

2 But what about negative Numbers ? Sincerhas to be positive (since it is a distance), using arg(z) = 0 only includes thepositive Numbers . From looking at figure 1, we can determine that we also needto include the possibility arg(z) = .The reason is that the function tan( ) is -periodic. So for anyn Z, wehavetan(arg(z)) = 0 arg(z) =n This means that arg(z) = 0 is only the solution forn= 0. Other valid solutionsinclude , 2 , .. Positive real Numbers are covered by even multiples of (including 0), and negative Numbers are covered by odd a Complex numberzwith magnituderand argument , whatareaandbsuch thatz=a+ib?One more flashback of trig: defining arg(z) andr |z|, we havea=rcos( )andb=rsin( ).Exercise that (a+ib)(a ib) =a2+b21 Just a little algebra:(a+ib)(a ib) =a2 iab+iab i2b2=a2+ 0 ( 1)b2=a2+b2 Exercise that (r ei )(r e i ) =r2 Remember that when you multiply exponentials, the exponent adds. So we have(r ei )(r e i ) =r2ei e i =r2ei i =r2e0=r2 Exercise is the Complex conjugate of a real number ?

3 For a real number , we can writez=a+ 0i=afor some real numbera. So thecomplex conjugatez =a 0i=a, which is also equal toz. Soarealnumberisitsowncomplexconjugate. [Suggestion : show this using Euler sz=r ei representation of Complex Numbers .]Exercise a point in the Complex plane. In the cartesian picture, howdoes the act of taking the Complex conjugate move the point? What about inthe polar coordinate picture?In the cartesian picture, we havea+ib, which becomesa ib. This is equivalentto taking the point (a, b) and moving it to (a, b), which is a reflection about thex-axis. SointheCartesianpicture, the polar coordinate picture, we change an angle from positive to nega-tive. Sointhepolarcoordinatepicture, should think about these two pictures, and convince yourself that theyare 9(Advanced).Prove Euler s formula. [Hint : what s the Taylorseries (or MacLaurin series, actually) ofex? So what if you replacexbyi (remembering thati2n= ( 1)n)? Now what are the series expansions forcos( ) and sin( )?]

4 ]We begin with the MacLaurin series ofex:ex= j=0xjj!2 Now we make the change of variablesx i :ei = j=0(i )jj!= j=0(i )2j(2j)!+(i )2j+1(2j+ 1)!= j=0i2j 2j(2j)!+i2j+1 2j+1(2j+ 1)!= j=0 1j 2j(2j)!+ii2j 2j+1(2j+ 1)!= j=0 1j 2j(2j)!+i( 1)j 2j+1(2j+ 1)!= j=0 1j 2j(2j)!+i j=0 1j 2j+1(2j+ 1)!=cos( ) +isin( )The first step splits the sum into even and odd terms. The rest is just manipula-tion until the last step indentifies the series expansions found as those of sin( )and cos( ).Strictly speaking, we have to prove that all these series converge over whatis called an infinite radius of convergence. But we ll leave that problem to thefolks who have taken Complex Euler s Formula, show that the simple rule for complexconjugation gives the same results in either real/imaginary form or magni-tude/argument form. [Hint: take a Complex numberz=rei and defineaandbsuch thatrei =a+ib. Then take the Complex conjugate.]Using the results from exercises ** and **, forzwe havea=rcos( ) andb= sin( ) in cartesian (real/imaginary) form.

5 Forz =r e i ,a=rcos( ) =rcos( ) andb=rsin( ) = rsin( ) in cartesian form. Comparing these, wehave thataforzequalsaforz andbforzequals bforz .Exercise other formula are often grouped in with Euler s are:cos( ) =12(ei +e i )andsin( ) =12i(ei e i )Prove these using Euler s formula as given in equation??. [Hint: sin( x) = sin(x) and cos( x) = cos(x).]3 The trick is to use Euler s formula twice. For the positive angle, we haveei = cos( ) +isin( )and for the negative angle, we havee i = cos( ) +isin( )= cos( ) isin( )where the second step comes from the parity (even/odd-ness) of the sin and cosfunctions, which was given in the all we have to do is either add or subtract the functions. If we addthem, we findei +e i = (cos( ) +isin( )) + (cos( ) isin( ))= 2 cos( )From that, we get12(ei +e i )= cos( ).On the other hand, if we subtract them, we findei e i = (cos( ) +isin( )) (cos( ) isin( ))= 2isin( )And from there we easily obtain12i(ei e i )= sin( ).Exercise 12(Advanced).

6 There s a famous formula in mathematics whichcombines several of the most important mathematical constants:e, ,i, and a formula which is equal to zero, using each of those constants oncein your expression. [Hint : remember that inei is in radians.]We ll go straight to the answer:ei + 1 = 0. A friend bought me a pin withthis formula on it. Convince yourself that it is is the square root ofi?Following the methodology outlined in the text, we first convertito Euler snotation. It has modulus 1 and argument /2. So i= ei /2 Now we use the fact that z=z1/2and we have i=(ei /2)1/2= ei /212= ei /4 Since 1 =ei( /2 /2), we have i=ei( /2 /2)ei /4=ei( /2 /2+ /4)=ei(3 /4 /2)=ei /4ore5 /44 Converting these back to real part/imaginary part notation:ei /4= cos( 4)+isin( 4)=1 2+i 2ande5i /4= cos(5 4)+isin(5 4)= 1 2 i 2 This exercise is part of an interesting subject in mathematics called thenthroots of unity. M&S give more detail in their exercise **.Exercise de Moivre s formula,(cos( ) +isin( ))n= cos(n ) +isin(n )where Randn N.

7 [Hint : (eb)c=ebc]Once we have Euler s formula, this is pretty straightforward. Thanks to Euler,we have(cos( ) +isin( ))n=(ei )n=e(i )n=ei( n)=cos(n ) +isin(n )This also reminds us of another important rule when dealing with exponentials,which was given in the 15(Advanced).The technique described above can be used to findmany trigonometric identities. By first taking the trig function, then using theformulae given by equations??and??, doing some math with the result, thenconverting them back to trigonemetric forms, you can rather easily obtain manyresults from trigonometry. As an example, trysin2( ) + cos2( ) = 1(To the real show-offs: try dxsin2(ax) cos2(ax) = 132asin(4ax) +x8)As the exercise suggests, we first replace the sin and cos functions, then do somearithmetic:sin2( ) + cos2( ) =(ei e i 2i)2+(ei +e i 2)2=1 4(ei e i )2+14(ei +e i )2=14( (ei e i )2+(ei +e i )2)=14( (e2i 2 +e 2i )+(e2i + 2 +e 2i ))=14( e2i + 2 e 2i +e2i + 2 +e 2i )=144 =15 Now, for the show-off version.

8 I ll do the math a little less explicitly, butyou should be able to at least follow the gist: dxsin2(ax) cos2(ax) = dx(12i(eiax e iax))2(12(eiax+e iax))2= dx 116(e2iax 2 +e 2iax)(e2iax+ 2 +e 2iax)= dx 116(e4iax 2 +e 4iax)= 116( dx e4iax dx2 + dx e 4iax)= 116(14iae4iax 2x+1 4iae 4iax)= 11614ia(e4iax e 4iax)+x8= 132asin(4ax) +x86