Math 1050-1
Homework Set 6, Problem 9 explanation
Problem 9 gives you a complicated complex number, and then asks you to compute powers of that
number. There are three methods to go about solving this problem. For this explanation we will use a
different number. In our problem we have:
√
1
3
i
z=− +
2
2
(1)
The Long, Tedious Method
This method involves simply solving the problem as it is given.
z
=
z2
=
=
=
z3
=
√
1
3
− +
i
2
2
√
√
1
1
3
3
(− +
i)(− +
i)
2√ 2
2
2
3
1
3
+
i−
4
2√
4
3
1
i
− +
2
2
√
√
1
1
3
3
(− +
i)(− +
i)
2
2
2
2
(2)
(3)
(4)
(5)
(6)
And so on. Blah.
The Less-Long, Less-Tedious Method
This method involves getting approximate values for the real and complex part of z, and then doing a
similar process as above. So we would have:
z
z
2
z3
√
3
1
= − +
i
2
2
≈ 0.5 + 0.866i
≈ (0.5 + 0.866i)(0.5 + 0.866i)
(7)
(8)
(9)
≈ 0.0025 + 0.866i − 0.749956
≈ −0.747 + 0.866i
(10)
(11)
≈ (−0.747 + 0.866i)(0.5 + 0.866i)
(12)
And so on.
The Snazzy Way
This method relies on a fact that you might not know. The fact is:
exi = cos x + i sin x
For your problem in WebWorks z = e
For our problem, z = e
2π
3
2π
5 i
2π
= cos( 2π
5 ) + i sin( 5 ) =
2π
1
= cos( 2π
3 ) + i sin( 3 ) = − 2 +
√
3
2 i.
√
−1+ 5
4
+
√ √ √
2 5+ 5
i.
4
So we have:
z
z
2
2π
3 i
= e
= (e
(13)
2
)
4π
3 i
= e
z3
2π
3 i
4π
4π
= cos( ) + i sin( )
3
3
2π
i 3
3
= (e )
6π
3 i
= e
2πi
(14)
(15)
(16)
(17)
(18)
= e
= cos(2π) + i sin(2π)
(19)
(20)
= 1
(21)
If you do not know trigonometric functions, that is fine. WebWork will accept your answers exactly as
cos(2*pi/3) + sin(2*pi/3) i, or whatever you happen to have.