3—Complex Algebra 72Examples
Simplify these expressions, making sure that you can do all of these manipulations yourself.
3 − 4 i
2 −i=
(3− 4 i)(2 +i)
(2−i)(2 +i)=
10 − 5 i
5= 2−i.(3i+ 1)^2[
1
2 −i+
3 i
2 +i]
= (−8 + 6i)[
(2 +i) + 3i(2−i)
(2−i)(2 +i)]
= (−8 + 6i)5 + 7i
5=
2 − 26 i
5.
i^3 +i^10 +i
i^2 +i^137 + 1=
(−i) + (−1) +i
(−1) + (i) + (1)=
− 1
i=i.Manipulate these using the polar form of the numbers, though in some cases you can do it either way.
√
i=(
eiπ/^2) 1 / 2
=eiπ/^4 =1 +i
√
2.
(
1 −i
1 +i) 3
=
(√
2 e−iπ/^4
√
2 eiπ/^4) 3
=
(
e−iπ/^2) 3
=e−^3 iπ/^2 =i.(
2 i
1 +i√
3
) 25
=
(
2 eiπ/^2
2( 1
2 +i1
2√
3
)
) 25
=
(
2 eiπ/^2
2 eiπ/^3) 25
=
(
eiπ/^6) 25
=eiπ(4+1/2)=iRoots of Unity
What is the cube root of one? One of course, but not so fast; there are three cube roots, and you can easily find
all of them using complex exponentials.
1 =e^2 kπi, so 11 /^3 =(
e^2 kπi) 1 / 3
=e^2 kπi/^3andkis any integer.k= 0, 1 , 2 give
11 /^3 = 1, e^2 πi/^3 = cos(2π/3) +isin(2π/3),=−1
2
+i√
3
2
e^4 πi/^3 = cos(4π/3) +isin(4π/3)=−1
2
−i