EXERCISE SHEET 8
I thank Inka Busack for pointing out a mistake in a previous version of these
solutions.
(1) Write, in simplified Cartesian form, all solutions to
(z − 1)3 = 8.
Solution: The equation
(z − 1)3 = 8
means that z − 1 is one of the three cube roots of 8: ζ0 (8), ζ1 (8), ζ2 (8) such
that
ζi3 = 8.
To determine the ζi , i = 0, 1, 2, we put 8 in polar form
8 = 8 · ei·0 .
Thus the cube roots of 8 are given by
√
3
ζ0 = 8 = 2
√
√
3
ζ1 = 8e2πi/3 = 2eiπ−iπ/3 = 2eiπ e−iπ/3 = −1 + 3i
√
√
3
ζ2 = 8e4πi/3 = 2eiπ eiπ/3 = −1 − 3i.
Setting z − 1 = ζi for i = 0, 1, 2, we find the three solutions
z1 = 3
√
z2 = 3i
√
z3 = − 3i.
(2) Find all solutions, in Cartesian form (a + ib) of
z 4 + 8iz = 0.
Solution: Begin by factoring:
z 4 + 8iz = z(z 3 + 8i).
This tells us that
z 4 + 8iz
if z = 0 or z 3 + 8i = 0. Thus one solution is z0 = 0. To find the others, we
put 8i in polar form:
8i = 8eiπ/2 .
The three cube roots of 8i are given by
√
√
3
ζ0 = 8eiπ/6 = 3 + i
√
√
3
ζ1 = 8eiπ/6+i2π/3 = − 3 − i
√
3
ζ2 = 8eiπ/6+i4π/3 = −2i.
1
2
EXERCISE SHEET 8
Each ζi is also
√ a solution of√our original equations. So the solutions are
z0 = 0, z1 = 3 + i, z2 = − 3 − i, and z3 = −2i.
(3) Find in Cartesian form all solutions of
(z 3 + i)2 + 4 = 0.
Solution: Rewrite this equation as
(z 3 + i)2 = −4.
This means that z 3 + i is a square root of −4, that is
z 3 + i = 2i
or
z 3 + i = −2i
. In the first case, we have
z 3 = i = eiπ/2 ,
so z is equal to one of the three cube roots of i:
√
3
1
iπ/6
i
z1 = e
= +
2
2√
3 1
− i
z2 = eiπ/6+2π/3 = −
2
2
iπ/6+4π/3
z3 = e
= −i.
In case
z 3 = −2i,
z is one of the three cube roots of
−2i = 2e−π/2 .
These are
√
!
3 1
z4 = 2e
= 2
− i
2
2
√
√
√
3
3
3
z5 = 2e−iπ/6+2π/3 = 2eiπ/2 = 2i
!
√
√
√
3 1
3
3
−iπ/6+4π/3
z6 = 2e
=− 2
+ i .
2
2
√
3
−iπ/6
√
3
The six numbers z1 , z2 , z3 , z4 , z5 and z6 above represent all solutions to
the equation (z 3 + i)2 + 4 = 0.
(4) Find all complex numbers z satisfying
z 6 = z 3 − 1.
Solution: Begin by completing the square:
2
1
3
6
3
3
z −z +1= z −
+ .
2
4
So the quantity
z 3 − 1/2
EXERCISE SHEET 8
3
is one of the square roots of − 34 :
√
1
3
z − =
i
2
2
or
√
1
3
z3 − = −
i,
2
2
that is:
√
3
1
3
z = +
i = eiπ/3
2
2
or
√
3
1
3
i = e−iπ/3 .
z = −
2
2
In the first case, we have the three roots:
3
z0 = eiπ/9
z1 = eiπ/9+2π/3 = ei7π/9
z2 = eiπ/9+4π/3 = ei13π/9 .
In the second case, we have the roots:
z3 = e−iπ/9
z4 = e−iπ/9+2π/3 = ei5π/9
z5 = e−iπ/9+4π/3 = ei11π/9 .
The numbers z0 , z1 , z2 , z3 , z4 , and z5 represent all solutions of the equation.
(5) Find the real part of
√
√
(cos( 2) + i sin( 2))137 .
Solution: By de Moivre’s formula:
√
√
√
√
(cos( 2) + i sin( 2))137 = cos( 2 · 137) + i sin( 2 · 137),
√
which has real part cos(137 2).