Complex Numbers — Practice
One new number, i, and every polynomial equation finally has a solution.
What these questions cover
These questions use 6 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Exponent — How many times the base is used as a factor. In 53 the exponent is 3.
- Binomial — A polynomial with two terms.
- Roots — The x values where y = 0.
- Discriminant — b2 - 4ac. Its sign counts the real solutions.
- Remainder — What is left when the divisor no longer fits. Always lower degree than the divisor.
- Conjugate — a - bi is the conjugate of a + bi. Multiplying them clears the i entirely.
Problem 1
Simplify i2.
Worked answer
-1
i2=-1
Problem 2
Simplify i3.
Worked answer
-i
i2 · i = -1 · i = -i
Problem 3
Simplify i12.
Worked answer
1
12 ÷ 4 leaves remainder 0
i0 pattern value is 1
Problem 4
Write √-25 in terms of i.
Worked answer
5i
√-25=√25√-1=5i
Problem 5
Simplify (3+2i)+(5-6i).
Worked answer
8-4i
3+5=8
2i-6i=-4i
Problem 6
Simplify i39.
Worked answer
-i
39 = 4(9)+3
i3=-i
Problem 7
Simplify (7-i)-(2+4i).
Worked answer
5-5i
7-2=5
-i-4i=-5i
Problem 8
Expand (2+i)(3-2i).
Worked answer
8-i
6-4i+3i-2i2
-2i2=+2
8-i
Problem 9
Expand (1+3i)2.
Worked answer
-8+6i
1+6i+9i2
9i2=-9
-8+6i
Problem 10
Give the conjugate of 4-7i.
Worked answer
4+7i
4+7i
Problem 11
Expand (5+2i)(5-2i).
Worked answer
29
25-4i2
-4i2=+4
29
Problem 12
Simplify 103-i.
Worked answer
3+i
Bottom: (3-i)(3+i)=9+1=10
Top: 10(3+i)=30+10i
30+10i10=3+i
Problem 13
Simplify 1+i1-i.
Worked answer
i
Bottom: (1-i)(1+i)=1+1=2
Top: (1+i)2=1+2i-1=2i
2i2=i
Problem 14
How many real roots does x2+2x+5=0 have?
- None — both roots are complex
- One
- Two
- Three
Worked answer
None — both roots are complex
4-20=-16
Negative discriminant means no real roots
Problem 15
Solve x2+9=0. Give the positive-imaginary root.
Worked answer
3i
x2=-9
x=± 3i
Problem 16
Simplify √-4 · √-9.
Worked answer
-6
2i · 3i = 6i2
=-6
Multiplying under the roots first would wrongly give +6