40
Example 3.2 Givena,b∈R. Findx,y∈Rexpressed by aandb, such that (x+iy)2=a+ib.
When we compute the left hand side, we get (x+iy)2=x2−y2+ 2ixy,
so by a splitting into the real and the imaginary part we obtain the two equations x2−y2=a og 2xy=b.
Thisimplies that a2+b2=
x2−y22
+ 4x2y2=
x2+y22 , hence
x2+y2=
a2+b2≥0.
When this is compared withx2−y2=a, we get x2= a+√
a2+b2
2 (≥0), y2= −a+√ a2+b2
2 (≥0),
thus (2) x=±
a+√
a2+b2
2 og y=±
−a+√ a2+b2
2 .
Hence a solution isnecessarilyof the form (2). We see, however, that (2) usually gives four possibilities, and they cannot all be solutions, because we know that there are only two solutions. Hence we must check all our possible solutions.
The equationx2−y2=ais of course always satisfied, so we turn towards 2xy=b.
If b = 0, then either x= 0 or y = 0, according to (2), and the equation 2xy =b = 0 is of course fulfilled. (In this case (2) produces actually only two solutions).
Ifb= 0, then a check shows that the solution is
x=±
a+√ a2+b2
2 og y=±b
|b|
−a+√ a2+b2
2 ,
where the signs are corresponding.
Example 3.3 Find all the cubic roots of (a) 1, (b) −8, (c) i.
(–1-i*sqrt(3))/2 (–1+i*sqrt(3))/2
1
Figure 13: (a) The cubic roots of 1.
(a) The three cubic roots are of course 1, −1
2 +i
√3
2 , −1 2−i
√3 2 , i.e.
e0, exp
i2π 3
, exp
i4π
3
.
1-i*sqrt(3) 1+i*sqrt(3)
–2
Figure 14: (a) The cubic roots of−8.
(b) The three cubic roots of−8 are
√ √
42 i.e.
2eiπ, 2 exp
iπ 3
, 2 exp −iπ
3
.
-i
(sqrt(3)+i)/2 (-sqrt(3)+i)/2
Figure 15: (a) The cubic roots ofi.
Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more
American online LIGS University
▶ enroll by September 30th, 2014 and
▶ save up to 16% on the tuition!
▶ pay in 10 installments / 2 years
▶ Interactive Online education
▶ visit www.ligsuniversity.com to find out more!
is currently enrolling in the Interactive Online BBA, MBA, MSc,
DBA and PhD programs:
Note: LIGS University is not accredited by any nationally recognized accrediting agency listed by the US Secretary of Education.
More info here.
(c) The three cubic roots ofiare
−i, −
√3 2 +i1
2,
√3 2 +i1
2, thus
exp −iπ
2
, exp
i5π
6
, exp
iπ
6
.
Example 3.4 Find all complex roots of (a) √
i, (b) √3
−1 +i, (c) √4
−1, (d) √5
−32.
1
Figure 16: (a) The square roots ofi.
(a) The symbol√
imeans the roots of the equation z2=i= exp
i
π 2 + 2pπ
, p∈Z, so
z= exp π
4 +pπ
=±exp
iπ 4
=± 1
√2(1 +i).
(b) The symbol √3
−1 +iis the set of roots of the equation z3=−1 +i=√
2 exp
i 3π
4 + 2pπ
, p∈Z, so
z=√6 2 exp
i
π
+p·2π
, p= 0,1,2,
44
2^{1/6)
Figure 17: (a) The cubic roots of −1 +i.
or, more explicitly, z1 = √6
2· 1
√2+i 1
√2
= 1
√3
2(1 +i), z2 = 1
√3
2(1 +i)·1
2(−1 +i√
3) = 1 2√3
2(−1−√
3 +i{−1 +√ 3}),
z3 = 1
√3
2(1 +i)·1
2(−1−i√
3) = 1 2√3
2(−1 +√
3−i{1 +√ 3}).
1
Figure 18: (a) The quadruple roots of−1.
(c) The symbol √4
−1 is the set of roots of the equation z4=−1 = exp(i(π+ 2pπ)), p∈Z,
so
z= exp iπ
4 +pπ 2
, p= 0,1,2, 3,
–2
Figure 19: (a) The fifth roots of −32.
or more explicitly,
z∈ 1
√2+i 1
√2,− 1
√2+i 1
√2,− 1
√2 −i 1
√2, 1
√2−i 1
√2
. (d) The symbol √5
−32 means the set of solutions of the equation z5=−32 = 25ei(π+2pπ), p∈Z,
the solution of which are z= 2 exp
i
π
5 +p·2π 5
, p= 0,1,2,3,4.
Remark 3.1 Note that we forp= 2 obtain the trivial solutionz=−2. The remaining rootscan be expressed by means of square roots, if we apply the results of the folloing Example 3.5. ♦
Example 3.5 Prove that cosπ
5 = 1 +√ 5
4 and sinπ
5 =
10−2√ 5
4 .
Hint: Apply that
cosπ
5 +isinπ 5
5
=−1, and putx= cosπ
5. If we put x= cosπ
5 >0, then sinπ 5 =√
1−x2>0, hence 0 =
cosπ
5 +isinπ 5
5 + 1 =
x+i
1−x2 5
+ 1
= x5+ 5i x4
1−x2−10x3 1−x2
−10i x2
1−x2 1−x2 22 22
46
When we split into the real and the imaginary part we get the two equations
(3)
⎧⎪
⎨
⎪⎩
0 =x5−10x3 1−x2
+ 5x
1−x22 + 1, 0 =√
1−x2·
5x4−10x2 1−x2
+
1−x22 . We have assumed that √
1−x2= 0, hence (3) is reduced to
⎧⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎩
0 = x5+ 10x5−10x3+ 5x5−10x3+ 5x+ 1 = 16x5−20x3+ 5x+ 1
= (x+ 1)
16x4−16x3−4x2+ 4x+ 1
0 = 5x4+ 10x4−10x2+x4−2x2+ 1 = 16x4−12x2+ 1,
and sincex=−1 is not a common solution, it follows that we shall only solve the following reduced system of equations,
⎧⎨
⎩
16x4 − 16x3 − 4x2 + 4x + 1 = 0,
16x4 − 12x2 + 1 = 0.
Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more
.
Then by a subtraction, 16x3−8x2−4x= 4x
4x2−2x−1
= 0.
The solutions are 0 and x= 2±√
4 + 16
8 =2±2√ 5
8 = 1±√ 5 4 ,
so the solution of the original equations must be one of these.
Since (3)has a solutionx= cosπ
5 >0, and since the only positive of the possible solution is x= 1 +√
5 4 , we conclude that
cosπ
5 = 1 +√ 5 4 , hence
sinπ 5 =
1−
1 +√ 5 4
2
=
16−1−6−2√ 5
4 =
10−2√ 5
4 .
Alternatively, it follows from (3) that in particular, 0 = 16x5−2 +x3+ 5x+ 1 = (x+ 1)
16x4−16x3−4x2+ 4x+ 1 . Since
16x4−16x3−4x2+ 4x+ 1 = 16x4−16x3−8x2+ 4x2+ 4x+ 1
= 4x22
−8x2(2x+ 1) + (2x+ 1)2=
4x2−2x−12 , it follows that x= cosπ
5 ∈]0,1[ fulfils the equation 16x5−20x3+ 5x+ 1 = (x+ 1)
4x2−2x−12
= 4(x+ 1)
x−2 +√ 4 + 16 8
2
x−2−√ 4 + 16 8
2
= 4{x−(−1)}
x−1 +√ 5 4
2
x−1−√ 5 4
2 , and we conclude thatx= cosπ
5 ∈]0,1[ belongs to the set
−1,1 +√ 5 4 ,1−√
5 4
.
Since only 1 +√ 5
4 is positive, we conclude π 1 +√
5
48 and then
sinπ 5 =
10−2√ 5 4 is found as above.
k_10 k_10
1-k_10 k_10
1
E
D C
B
A
Alternatively, the example can be solved geometrically by noting thatABC is the same angle as BCD. Then
|AB|
|BC| = |BC|
|CD|, thus 1
k10 = k10 1−k10. We obtain the equation of second degree
k210+k10−1 = 0, hence
k10=−1 2
(−)+
!1 4 + 1 =
√5−1 2 ,
where we have exploited that k10>0. Finally, since|AB|= 1, cosπ
5 = |AE|=|AD|+1
2|DC|=k10+1
2|DC|=k10+1
2 (1−k10)
= 1
2 (1 +k10) = 1 +√ 5 4 .
Remark 3.2 The notation k10 is due to the fact that it is the length of the cord of the regular decagon, inscribed in the unit circle. ♦
Example 3.6 Find all roots of the equation z4+i= 0.
We rewrite this equation as z4=−i= exp
i
−π 2 + 2pπ
, p∈Z, thus
z= exp
i −π
8 +p· π 2
, p= 0,1,2, 3.
It follows from cosπ
8 =
!cosπ4+ 1
2 =
1 + √1
2
2 =
√ 2 + 1 2√
2 =
2 +√ 2
2 ,
and sinπ
8 =
!1−cosπ4
2 =
1−√12
2 =
√ 2−1 2√
2 =
2−√
2
2 ,
that
z1 = cosπ
8 −isinπ
8 = 1
2
2 +√ 2−i
2−√ 2
,
z2 = i z1 = 1
2
2−√ 2 +i
2 +√ 2
,
z3 = −z1 = 1
2 −
2 +√ 2 +i
2−√ 2
,
z4 = −i z1 = 1
2 −
2−√ 2−i
2 +√ 2
.
Example 3.7 Compute (−3 + 4i)−32.
It follows by inspection that
−3 + 4i= 1−4 + 2·2i= (1 + 2i)2, thus
(−3 + 4i)12 =±(1 + 2i), and hence
(−3 + 4i)−32 = 1
(−3 + 4i)(−3 + 4i)12 =∓ 1
(−3 + 4i)(1 + 2i) =∓(−3−4i)(1−2i) 25·5
= ±(3 + 4i)(1−2i)
125 =±3 + 8−6i+ 4i
125 =±11−2i 125 .
50 Example 3.8 Assume thatn∈N\ {1}. Prove that
sinπ
n ·sin2π
n · · ·sin(n−1)π
n = n
2n−1.
Hint: Prove that the left hand side can be written as 1
2n−1 times the product of the roots of the polynomial (1−z)n−1, which are different from zero.
The equation (1−z)n−1 = 0 has the solutions 1−z= exp
i2pπ
n
, p= 0,1, . . . , n−1, so when we rewrite each solution in the following way,
zp = 1−exp
i2pπ n
= 1−cos2π
n −isin2pπ
n = 1−
1−2 sin2pπ n
−i·2 cospπ
n ·sinpπ n
= 2 sinpπ n ·
sinpπ
n −icospπ n
= 2 sinpπ n ·
1 i exp
ipπ
n
,
Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more
www.mastersopenday.nl Visit us and find out why we are the best!
Master’s Open Day: 22 February 2014
Join the best at
the Maastricht University School of Business and Economics!
Top master’s programmes
• 33rd place Financial Times worldwide ranking: MSc International Business
• 1st place: MSc International Business
• 1st place: MSc Financial Economics
• 2nd place: MSc Management of Learning
• 2nd place: MSc Economics
• 2nd place: MSc Econometrics and Operations Research
• 2nd place: MSc Global Supply Chain Management and Change
Sources: Keuzegids Master ranking 2013; Elsevier ‘Beste Studies’ ranking 2012;
Financial Times Global Masters in Management ranking 2012
Maastricht University is the best specialist
university in the Netherlands
(Elsevier)
then sinπ
n ·sin2π
n · · ·sin(n−1)π
n = in−1 2n−1
n−1"
p=1
exp −ipπ
n n−1"
p=1
zp
= in−1 2n−1 exp
−iπ n· 1
2(n−1)n n−1"
p=1
zp= 1 2n−1 exp
(n−1)iπ 2
exp
−(n−1)iπ 2
n−1"
p=1
zp
= 1
2n−1
n−1"
p=1
zp. Now, we also have
(1−z)n−1 = n j=0
n j
(−1)jzj−1 = n j=1
n j
(−1)jzj =−z
n−1"
p=1
(zp−z)
= −n−1"
p=1
zp·z+· · ·+ (−z)n. so by identifying the coefficients for j= 1 we get
−n−1"
p=1
zp= n
1
(−1)1=−n,
hence by insertion, sinπ
n ·sin2π
n · · ·sin(n−1)π
n = n
2n−1.
Example 3.9 Solve the equationzn=z for everyn∈N.
Also, solve the equation, whenn∈Z.
If n = 1, then the equation becomes z = z. The set of complex numbers which are equal to their complex conjugated, is equal to the set of real number, hence the solution isR.
Ifn≥2, thenz= 0 is trivially a solution.
Then assume thatz= 0. Putz=r ei θ,r >0, which gives rneinθ =r e−iθ, r >0,
thus
rn−1ei(n+1)θ= 1.
We conclude that r= 1 andθ= 2pπ
n+ 1,p= 0, 1,. . .,n, so the equation has then+ 2 solutions z = 0 and z = exp
i 2pπ
, p= 1,2, . . . , n+ 1.
52
Ifn= 0, then the equation is reduced toz= 1, and it follows that the solution isz= 1.
Finally, let n <0. If we write n=−m, m∈N, thenz−m=z. In particular, 0 is never a solution.
Ifm= 1, thenz−1=zis rewritten as 1 =z·z=|z|2,
and the set of solutions is the unit circle.
Whenm >1, we use polar coordinatesz=r eiθ, so 1 =z zm=rm+1·e−iθ·eimθ=rm+1ei(m−1)θ, hence r= 1 andθ= 2pπ
m−1. Thus we obtain in this casem−1 =|n| −1 =−n−1 solutions, zp= exp
i 2pπ
m−1
= exp
−i 2pπ n+ 1
, p= 1,2, . . . , m−1 =−n−1.
Example 3.10 Prove that the function
f(z) = 2x4−12x2y2+ 2y4−3x3+ 9xy2+i
8x3y−8xy3−9x2y+ 3y3 , can be written as a polynomial in the complex variable z.
Then find the roots.
Concerning polynomials, a good strategy is to identify the degrees of the pair (x, y), which occur. We see that we have the degrees 4 and 3, and since
z4= (x+iy)4=x4+ 4ix3y−6x2y2−4ixy3+y4=x4−6x2y2+y4+i
4x3y−4xy3 and
z3= (x+iy)3=x3+ 3ix2y−3xy2−iy3=x3−3xy2+i
3x2y−y3 , it follows that
f(z) = 2x4−12x2y2+ 2y4−3x3+ 9xy2+i
8x3y−8xy3−9x2y+ 3y3
= 2
x4−6x2y2+y4+i
4x3y−4xy3
−3
x3−3xy2+i
3x2y−y3
= 2z4−3z3, thus
f(z) = 2z4−3z3= 2z3
z−3 2
,
and the roots arez= 0 (of multiplicity 3) and the simple rootz= 3 2.
Example 3.11 (Enestr¨om’s theorem). Given the real numbersp0,p1,. . .,pn, for which p0> p1> p2>· · ·> pn>0.
Prove that the polynomial
P(z) =p0+p1z+p2z2+· · ·+pnzn
does not have a zero in the open unit disc|z|<1.
Hint: Consider(1−z)P(z).
When we compute (1−z)P(z) we obtain (1−z)P(z) =
p0+p1z+p2z2+· · ·+pnzn
−
p0z+p1z2+· · ·+pn−1zn+pnzn+1
= p0−(p0−p1)z−(p1−p2)z2− · · · −(pn−1−pn)zn−pnzn+1. According to we assumption we havepj−1−pj >0, so when|z|<1 we get the estimate
|(1−z)P(z)|> p0−(p0−p1)−(p1−p2)− · · · −(pn−1−pn)−pn= 0,
and we conclude that (1−z)P(z)= 0 for |z|<1, so in particular thatP(z) does not have any zero in the open unit disc.
> Apply now
redefine your future
AxA globAl grAduAte
progrAm 2015
54