Complex Numbers — Challenge
One new number, i, and every polynomial equation finally has a solution.
What these problems 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
What is i2?
- -1
- 1
- i
- -i
- 0
Worked answer
-1
The imaginary unit i is defined as a number whose square is -1.
So i2 = -1 by definition.
This is the whole point of introducing i: it gives -1 a square root.
Problem 2
What is (2+3i)+(1-i)?
- 3+4i
- 3+2i
- 1+2i
- 3-2i
- 2+2i
Worked answer
3+2i
Add the real parts and the imaginary parts separately.
Real: 2+1 = 3. Imaginary: 3i-i = 2i.
So the sum is 3+2i.
Problem 3
What is i3?
- i
- -i
- 1
- -1
- 0
Worked answer
-i
Write i3 as i2· i.
Since i2 = -1, this is -1· i = -i.
The powers of i cycle through i, -1, -i, 1 and then repeat.
Problem 4
What is (1+i)(1-i)?
- 0
- 1
- 2
- 2i
- -2
Worked answer
2
Expanding gives 1-i+i-i2, and the two middle terms cancel.
That leaves 1-i2, and since i2 = -1 this is 1-(-1).
So the answer is 2, a real number: multiplying a complex number by its conjugate always gives a real result.
Problem 5
What is the conjugate of 3-4i?
- -3+4i
- 3+4i
- -3-4i
- 4-3i
- 3-4i
Worked answer
3+4i
The conjugate keeps the real part and changes the sign of the imaginary part.
So 3-4i becomes 3+4i.
Negating both parts would give -3+4i, which is the negative rather than the conjugate.
Problem 6
What is |3+4i|?
- 3
- 4
- 5
- 7
- 25
Worked answer
5
The modulus is the distance from the origin on the complex plane, found by Pythagoras.
√32+42 = √25 = 5.
Adding the parts gives 7 and forgetting the square root gives 25; both appear among the options.
Problem 7
What is i4?
- 1
- -1
- i
- -i
- 0
Worked answer
1
Write i4 as i2· i2.
That is (-1)(-1) = 1.
Because i4 = 1, the powers of i repeat with period four, so i5 = i and so on.
Problem 8
If z=1+i, what is z4?
- -4
- 4
- 4i
- -4i
- 0
Worked answer
-4
z2=(1+i)2=1+2i+i2=2i. Then z4=(z2)2=(2i)2=4i2=-4. Squaring twice is far less work than expanding a fourth power.