35
Second variant. An unconscious application of the solution formula from high school gives the awkward solutions
z=−1±√
1 + 2 + 4i=−1±√ 3 + 4i.
We assume that√
3 + 4i=a+ibfor somea,b∈R. Then by squaring, a2−b2+i2ab= 3 + 4i, i.e. a2−b2= 3, 2ab= 4,
thus
(a2+b2)2= (a2−b2)2+ (2ab)2= 9 + 16 = 25,
from whicha2+b2= 5. Together witha2−b2= 3 this givesa2= 4 and b2= 1. Now,ab >0, soaandbhave the same sign, i.e.√
3 + 4i=±(2 +i). Then by insertion z=−1±(2 +i) =
−3−i, 1 +i
C. The solutions ofnormed equations of second degree are checked by using that the sum of the roots is the coefficient ofzwith opposite sign, and the product of the roots is equal to the constant term.
In the present case we get (−3−i) + (1 +i) =−2,
(−3−i)(1 +i) =−3 + 1 +i(−3−1) =−2−4i.
Q.E.D.
Example 3.3 Find in the formz=a+ib,a,b∈R, the solutions of the equation z2−(5 + 5i)z+ 13i= 0.
A. Complex equation of second degree.
D. Apply the solution formula from high school in its complex form.
I. We get by the solution formula,
z = 1
2
5 + 5i± 52(1 +i)2−4·13i
= 1
2
5 + 5i± 52·2i−52i
= 1
2
5 + 5i±√
−2i = 1 2
5 + 5i± (1−i)2
=
⎧⎪
⎪⎨
⎪⎪
⎩ 1
2(5 + 5i+ 1−i) = 3 + 2i, 1
2(5 + 5i−1 +i) = 2 + 3i.
C. The sum of the roots is (3 + 2i) + (2 + 3i) = 5 + 5i,
i.e. the coefficient of zof the opposite sign, The product of the roots is
(3 + 2i)·(2 + 3i) = 6−5 + 4i+ 9i= 13i, i.e. equal to the constant term. Q.E.D.
Example 3.4 Find in the formz=a+ib,a,b∈R, the solutions of the equation iz2−(2 + 3i)z+ 1 + 5i= 0.
A. A non-normed complex equation of second degree.
D. First multiply by−i; then solve the equation.
I. When we multiply by−iwe obtain the equivalentnormed equation (2) z2−(3−2i)z+ 5−i= 0.
Then by the usual solution formula known from high school,
z = 1
2
3−2i± (3−2i)2−4(5−i)
= 1
2
3−2i±√
9−4−12i−20 + 4i
= 1
2
3−2i±√
−15−8i
= 1
2
3−2i± (4i)2+ 1−2·4i·1
= 1
2
3−2i± (1−4i)2 = 1
2 {3−2i±(1−4i)}
=
2−3i, 1 +i.
C. The sum of the roots is (2−3i) + (1 +i) = 3−2i,
which is the coefficient ofz of the opposite sign in thenormed equation (2).
The product of the roots is
(2−3i)·(1 +i) = 2 + 3−3i+ 2i= 5−i, which is equal to the constant term in (2).
Thus, the solutions are correct for (2). Finally, when (2) is multiplied byi, we obtain the equivalent original equation.
37 Example 3.5 Find the solutions of the equations 1) (z+ 1)2= 3 + 4i.
2) (z+ 1)4= 3 + 4i.
A. Two disguised binomial equations which can be solved by taking a square root.
D. Solve the equations, where the second equations can be derived from the first one.
I. 1) From
(z+ 1)2= 3 + 4i= 4−1 + 2·2·i= (2 +i)2, we getz+ 1 =±(2 +i), hence
z=
1 +i,
−3−i.
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2) As above, (z+ 1)4= (2 +i)2, hence by (1), (z+ 1)2=
2 +i,
−2−i.
The four roots are z=−1±√
2 +i, og z=−1±√
−2−i=−1±i√ 2 +i.
Remark. Let√
2 +i=a+ib, whereaandb∈R. Then we get by a squaring, a2−b2= 2 and 2ab= 1,
from which
(a2+b2)2= (a2−b2)2+ (2ab)2= 5, i.e.
a2+b2=√
5 and from previously a2−b2= 2. This gives
a=±
!√ 5 + 2
2 , b=±
!√ 5−2
2 , ab >0, hence,
√2 +i=±
⎧⎨
⎩
!√ 5 + 2
2 +i
!√ 5−2
2
⎫⎬
⎭. We can therefore write the four roots,
−1 +
!√ 5 + 2
2 +i
!√ 5−2
2 , −1−
!√ 5 + 2
2 −i
!√ 5−2
2 ,
−1 +
!√ 5−2
2 −i
!√ 5 + 2
2 , −1−
!√ 5−2
2 +i
!√ 5 + 2
2 . ♦
39 Example 3.6 Let ±√
w denote the two solutions of the equation z2 = w. Find the values of the following expressions written in the forma+ib:
1) ±√ 1 +i, 2) ± 1±√
i, 3) ± ±√
i.
A. Splitting into real and imaginary part by a square root. A very difficult example!
D. Letz2 =wfor given w. Insertz=a+iband find two equations for aandb. Then find in each case the corresponding equation without the symbols±√
·.
I. First some general theory.
Let z2 = w = u+iv, where u and v are given real numbers. If we put z = a+ib, we get by insertion and separation into real and imaginary parts that
a2−b2=u og 2ab=v.
Then
(a2+b2)2= (a2−b2)2+{2ab}2=u2+v2, and thus
a2+b2= u2+v2=|w|.
Since we already have thata2−b2=u, we get a2= |w|+u
2 , b2= |w| −u
2 , 2ab=v,
and we can findz =a+ib. The latter equation is used to check whetheraand b have the same or opposite sign.
1) z=±√ 1 +i.
Here, z2= (a+ib)2= 1 +i, thus u= 1 and|w|=√
2, andab >0. It follows from the above that
a=±
!√ 2 + 1
2 , b=±
!√ 2−1
2 , ab >0, hence
±√
1 +i=±
⎧⎨
⎩
!√ 2 + 1
2 +i
!√ 2−1
2
⎫⎬
⎭. C.Test. We get by a squaring
⎧⎨
⎩
!√ 2 + 1
2 +i
!√ 2−1
2
⎫⎬
⎭
2
=
√2 + 1
2 −
√2−1 2 + 2i
!√ 2 + 1
2 ·
√2−1
2 = 1 +i. Q.E.D.
2) z=± 1±√ i. Let us first find±√
i. Sincei= (1)π
2, it should now be well-known that
±√ i=±
1
√2 +i 1
√2
=± √
2 2 +
√2 2 i
.
For the first one of the two possibilities of 1±√
iwe get a)
w= 1 +
√2 2 +i
√2
2 , |w|=
1 +1 2 +√
2 + 1 2 =
2 +√
2,
u= 1 +
√2
2 =2 +√ 2
2 , 2ab=
√2 2 >0, a2=1
2
2 +√ 2 + 1
2(2 +√ 2)
= 1 4
2
2 +√
2 + (2 +√ 2)
, b2= 1
4
2
2 +√
2−(2 +√ 2)
, thus
z=±
1 2
2
2 +√
2 + (2 +√ 2) + i
2
2
2 +√
2−(2 +√ 2)
.
b) For the second one we have instead w= 1−
√2 2 −i
√2
2 , |w|=
1 +1 2 −√
2 + 1 2 =
2−√
2,
u= 1−
√2
2 =2−√ 2
2 , 2ab=−
√2 2 <0, a2=1
2
2−√ 2 + 1
2(2−√ 2)
= 1 4
2
2−√
2 + (2−√ 2)
, b2= 1
4
2
2−√
2−(2−√ 2)
, thus
z=±
1 2
2
2−√
2 + (2−√ 2)− i
2
2
2−√
2−(2−√ 2)
.
41 Hence the four roots are
1 2
2
2 +√
2 + (2 +√ 2) + i
2
2
2 +√
2−(2 +√ 2),
−1 2
2
2 +√
2 + (2 +√ 2)− i
2
2
2 +√
2(2 +√ 2), 1
2
2
2−√
2 + (2−√ 2)−i
2
2
2−√
2−(2−√ 2),
−1 2
2
2−√
2 + (2−√ 2) + i
2
2
2−√
2−(2−√ 2).
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.
C. It is possible to perform the test by noting that z =± 1±√
i are the four roots of the equation (z2−1)2=i.
a) If
z=±1 2
2
2 +√
2 + (2 +√ 2) +i
2
2 +√
2−(2 +√ 2)
, then
z2 = 1 4
2(2 +√ 2) + 2i
4(2 +√
2 =)−(2 +√ 2)2
= 1 +1 2
√ 2 +i
8 + 4√
2−4−2−4√ 2
= 1 + 1
√2(1 +i), hence
(z2−1)2=1
2(1 +i)2= 1
2·2i=i. Q.E.D.
b) If
z=±,1 2
2
2−√
2 + (2−√ 2)−i
2
2−√
2−(2−√ 2)
, then
z2 = 1 4
2(2−√ 2)−2i
4(2−√
2)−(2−√ 2)2
= 1 +1 2
−√ 2−i
8−4√
2−4−2 + 4√ 2
= 1− 1
√2(1 +i), hence
(z2−1)2=1
2(1 +i)2=i. Q.E.D.
3) z=± ±√ i.
In this case,z2=±√ i, i.e.
z4=i= (1)π
2+2pπ, p∈Z, and
z= (1)π
8+p·pi2, p= 0, 1,2,3,
43
where we have solve the equation as a binomial equation. Here, (1)π8 = cosπ
8 +isinπ 8
=
1 + cosπ4
2 +i
1−cosπ4 2
= 1
2
2 +√ 2 + i
2
2−√ 2, (1)π+π2 = (1)π8 ·i=−1
2
2−√ 2 + i
2
2 +√ 2, (1)π8+π = −(1)π8 =−1
2
1 +√ 2− i
2
2−√ 2, (1)π
8+3π2 = (1)π
8 ·(−i) =1 2
2−√
2− i 2
2 +√
2. All things put together we see that for± ±√
iwe get
±1 2
2 +√
2 +i
2−√ 2
, ±1
2
2−√ 2−i
2 +√
2
.
Example 3.7 Solve the equation of second degree,z2−4iz−1 + 4i= 0. Then find the roots of the polynomial
P(z) =z4−4iz2−1 + 4i.
A. An equation of second degree and another one in disguise.
D. Solve the equations, either by means of a formula or by inspection.
I. 1) We get by inspection,
z2−4iz−1 + 4i= (z2−1)−4i(z−1) = (z−1)(z+ 1−4i), so the first polynomial has the roots−1 + 4iand 1.
2) When we here replace z2 by z, we obtain the equation in (1). We therefore conclude that P(z) = 0, if and only if either
(i) z2= 1, or (ii) z2=−1 + 4i.
a) Whenz2= 1, the roots are±1.
b) Then we solvez2=−1 + 4i. If we putz=a+ib, we get a2−b2=−1, 2ab= 4, a2+b2=√
17, hence
a=±
!√ 17−1
2 =±1 2
2√
17−2,
b=±
!√ 17 + 1
2 =±1 2
2√
17 + 2.
Sincea·b >0, the roots are
±1 2
2√
17−2 +i
2√ 17 + 2
. C. Test.
1) Ifz=±1, thenP(±1) = 1−4i−1 + 4i= 0. Q.E.D.
2) If
z=±1 2
2√
17−2 +i
2√ 17 + 2
, then
z2 = 1 4
2·(−2) + 2i√
4·17−4
= 1
4{−4 + 2i·4·2}=−1 + 4i, hence by insertions,
P(z) = (−1 + 4i)2−4i(−1 + 4i)−1 + 4i
= (−1 + 4i)· {−1 + 4i+ 1−4i}= 0. Q.E.D.
Example 3.8 Find the solutionsz of the equation z2−z+ 1−i= 0.
Then find all complex solutions y of the equation e2y−ey+ 1−i= 0.
A. Complex roots.
D. One may either use the solution formula, or find a root by inspection.
I. 1) By a small inspection we see that
0 =z2−z+ 1−i= (z2+ 1)−(z+i) = (z+i)(z−1−i), thus the roots are 1 +iand−i.
2) If we instead apply the solution formula, we get
z = 1
2± 1
4−1 +i= 1 2 ±1
2
√−3 + 4i
= 1
2±1 2
√−4 + 1 + 2·2i·1 = 1
2 ± (1 + 2i)2
= 1
2{1±(1 + 2i)}=
1 +i,
−i.
45
3) If we putey =z, we get the same equation as before, so either ey= 1 +i= exp
ln 2 +i
π
4 + 2pπ , or
ey=−i= exp i
−π
2 + 2pπ .
From this we immediately get the solutions, either y= ln 2 +i
π
4 + 2pπ , p∈Z, or
y=i −π
2 + 2pπ , p∈Z.
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Example 3.9 1) Solve the complex equation z2−(1 + 2i)z+ 1−5i= 0.
2) Is it possible to write some of the solutions in the form z=√
2eit, t∈R?
A. A complex equation of second degree.
D. Use the solution formula, known from high school. Find modulus and argument of the solutions and compare.
I. 1) First we find the discriminant
B2−4AC = {−(1 + 2i)}2−4(1−5i)
= 1−4 + 4i−4 + 20i=−7 + 24i
= 32−42+ 2·3·4i= (3 + 4i)2. Then we get the solutions
z=1
2{1 + 2i±(3 + 4i)}=
2 + 3i,
−1−i.
2) Since|2 + 3i|=√
4 + 9 =√ 13=√
2, this solution is not possible.
Concerning the second solution we get
−1−i=√ 2·exp
−i3π 4 + 2ipπ
, p∈Z, thus
z=−1−i=√ 2·eit for
t=−3π
4 + 2pπ, p∈Z.
47