Easy Complex Numbers:
i2 = −1
Complex numbers are easy!
Just treat the imaginary number i like x and do regular algebra.
Whenever you see i2 you replace it: i2 = −1
Example :
(1 + i)(2 + i) = 2 + 3i + i2 = 2 + 3i − 1 = 1 + 3i
Good answers:
i
2+i
7 − 32 i
1.
1 + 2i + 3i2 =
2.
(1 + i)(1 + i) =
3.
(3 + i)(3 − i) =
4.
i + 2i + 3i + 4i =
5.
i + i3 =
Bad answers:
2i2
5/i
2 + 3i + i2
6.
1 + i + i2 + i3 =
7.
(i2 )(i2 ) =
8.
i4 =
9.
i5 =
Since i2 = −1 we have
10.
11.
12.
13.
√
√
√
√
−16 =
−25 =
−π 2 =
−π =
√
−1 = i, so
√
−9 =
√
√
−1 9 = 3i
The Power of I !!!!
(well . . . powers of i)
0.
i0 = 1
True for any number: A0 = 1
1.
i1 = i
True for any number: A1 = A
2.
i2 = −1
The definition of the imaginary number i
3.
i3 = i2 · i = −i
4.
i4 = i2 · i2 = (−1)(−1) = 1
-or-
(except A = 0)
Follows from i2 = −1
Follows from i2 = −1
i4 = i3 · i = (−i)(i) = −i2 = 1
Follows from i3 = −i
In this way one can reduce all the powers of i to 1, i, −1, or − i
Try doing this by completing this tableau:
i0 =
i4 =
i8 =
i12 =
i1 =
i5 =
i9 =
i13 =
i2 =
i6 =
i10 =
i14 =
i3 =
i7 =
i11 =
i15 =
i4 =
i8 =
i12 =
i16 =
Do you notice a pattern? How do you calculate i2019?
√
Roots of Negative Numbers:
0.
1.
2.
3.
4.
5.
6.
7.
8.
√
√
√
√
√
√
√
√
√
−1 = i
−4 =
−9 =
√
√
−1 = i
This is the basis for everything.
√
√
−1 4 = 2i
Factor out
−1, then solve.
√
−1 9 =
Solve this yourself.
−16 =
−25 =
−49 =
−100 =
−0.01 =
−π 2 =
π is the ratio of a circle’s circumference to its diameter
9.
√
−b2 =
b is a positive real number
10.
√
−π =
Algebra with i:
Just replace i2 with −1
0.
(1 + i)(2 + i) = 1 + 2i + i2 = 1 + 2i − 1 = 2i
i2 → −1
1.
3i(5 + 7i) = 15i + 21i2 = 15i − 21
i2 → −1
Note: this is the same as −21 + 15i
2.
(2 + 3i)(5 + 7i) =
3.
(2 + 3i)2 = (2 + 3i)(2 + 3i) =
4.
(2 + 3i)(2 − 3i) =
5.
(1 + i)(1 − i) =
6.
(3 + 5i)(3 − 5i) =
7.
(a + bi)(a − bi) =
a and b are real numbers.
a − bi is called the complex conjugate of a + bi
The product of a complex number and its conjugate is always real.
Complex Numbers:
Powers of i:
i0 = 1
i1 = i
Level 1
i2 = −1
fill in the rest below:
i0 =
i4 =
i8 =
i1 =
i5 =
i9 =
i2 =
i6 =
i10 =
i3 =
i7 =
i11 =
Notice the pattern ... it repeats every 4th power.
In general, i4n = 1 for any integer n.
√
√
−1 = i
−9 = 3i
Square Roots of Negative Numbers:
√
√
√
The trick is
−25 = −1 ∗ 25 = i ∗ 5 = 5i
1.
2.
3.
4.
5.
6.
√
√
√
√
√
√
−4 =
−16 =
−36 =
−49 =
−81 =
−144 =
Complex Numbers:
Algebra of i:
Level 2
treat i like a variable x, but always replace
Do algebra first, then replace i2 with −1 and then simplify.
1.
i(1 + i) =
2.
i(1 − i) =
3.
(1 + i)(1 + i) =
4.
(1 + i)(1 − i) =
5.
(2 + 3i)(2 − 3i) =
6.
(5 + 7i)(5 − 7i) =
7.
(a + bi)(a − bi) =
i2 → −1 :
Complex Numbers:
Powers of i
Level 3
we saw they repeat every 4th power:
i4n = 1
i4n+1 = i
i4n+2 = −1
i4n+3 = −i
So any power of i is determined by the remainder after dividing the power by 4:
i23 = i4∗5+3 = i4∗5i3 = (i4)5i3 = (1)5i3 = i3 = −i
23 = 4 ∗ 5 + 3
or
23
mod 4 = 3
or
23%4 = 3
The remainder of 23 after dividing the by 4 is 3, so i23 = i3 = −i
Find each of the following large powers of i:
1.
i17 =
2.
i31 =
3.
i54 =
4.
i68 =
5.
i131 =
6.
i231 =
7.
i931 =
Complex Numbers:
Level 4
Division and the Complex Conjugate
A complex number z = 2 + 3i has conjugate z ∗ = 2 − 3i (just negate the i part).
The product zz ∗ is always both real and positive:
zz ∗ = (2 + 3i)(2 − 3i) = 4 − 9i2 = 4 + 9 = 13
In general, if z = x + iy, the conjugate is z ∗ = x − iy and
zz ∗ = (x + iy)(x − iy) = x2 − i2 y 2 = x2 + y 2
This is useful in division, where we agree with the French and the Japanese that all denominators
should be real and positive. So whenever we see a complex denominator we multiply top and
bottom by its conjugate:
2 + 3i
(2 + 3i) (5 − 7i)
FOIL
31 + i
31
1
=
= 2
=
=
+ i
2
5 + 7i
(5 + 7i) (5 − 7i)
5 +7
74
74 74
Note the deceptively simple case with just i in the denominator:
(3 + 4i) (−i)
FOIL
4 − 3i
3 + 4i
=
=
=
= 4 − 3i
2
i
(i) (−i)
−i
1
Do division by multiplying top and bottom by the conjugate of the bottom:
1.
1+i
=
1−i
2.
2 + 5i
=
2 + 3i
3.
2 − 5i
=
2 − 3i
Complex Numbers:
Mixed Problems
1.
1 + i + i2 + i3 + i4 + i5 + i6 + i7 =
2.
1
1
+
=
1−i 1+i
3.
1 + i2 + i 4
=
i + i3 + i5
4.
5.
6.
√
−49 −
√
−36 =
(1 + i)4 = (1 + i)2 (1 + i)2 =
√3 + i 2
2
=
Level 5
Complex Numbers:
Level 6
Hard Problems
√3 + i 3
1.
=
2
2.
Suppose z = x + iy =
√
i
Find x and y by noting that z 2 = i