The value of (1 + i)6 is
- Square first: (1 + i)2 = 1 + 2i + i2 = 1 + 2i − 1 = 2i.
- Then (1 + i)6 = (2i)3 = 8i3.
- Since i3 = −i, this gives 8 × (−i) = −8i.
Answer: −8i
Complex numbers
NUST NET (Engineering) · Maths · Numbers, sets and functions. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.
The idea
A complex number is z = a + bi, where a is the real part, b is the imaginary part and i2 = −1. Add and subtract by collecting real and imaginary parts. Multiply like brackets in algebra, then replace i2 by −1.
Powers of i repeat in a cycle of four: i, −1, −i, 1. So for any power, only the remainder after dividing the index by 4 matters.
The conjugate z̄ = a − bi is the tool for division, because z × z̄ = a2 + b2 is real. The modulus |z| is the distance of the point (a, b) from the origin, and the argument is the angle its line makes with the positive real axis.
Rules to remember
- i2 = −1, i3 = −i, i4 = 1; for in, divide n by 4 and use the remainder as the new power
- Any four consecutive powers of i add up to 0
- Conjugate of z = a + bi is z̄ = a − bi; z + z̄ = 2a, z − z̄ = 2bi, z × z̄ = a2 + b2
- To divide, multiply top and bottom by the conjugate of the bottom
- Modulus: |z| = √(a2 + b2)
- |z1z2| = |z1| × |z2| and |z1/z2| = |z1| ÷ |z2|, so a modulus can be found without simplifying first
- Argument θ: tan θ = b/a, with the quadrant taken from the signs of a and b; principal value −π < θ ≤ π
Common mistake
Taking the argument straight from tan−1(b/a) without checking the quadrant. For z = −1 + i, tan−1(−1) gives −π/4, but the point lies in the second quadrant, so the argument is 3π/4.
Worked example
Practice questions
Try each one, then open the answer.
1. The principal argument of −1 + √3 i is
- A−π/3
- Bπ/3
- C2π/3
- D−2π/3
Show answer
Answer: C. The point (−1, √3) is in the second quadrant with reference angle tan−1√3 = π/3, so the argument is π − π/3 = 2π/3. Writing tan−1(−√3) = −π/3 directly ignores the quadrant.
2. The value of (1 − 2i)2 is
- A5 − 4i
- B−3 + 4i
- C−3 − 4i
- D−3
Show answer
Answer: C. (1 − 2i)2 = 1 − 4i + 4i2 = 1 − 4i − 4 = −3 − 4i. Taking i2 = +1 gives 5 − 4i, and dropping the middle term gives −3.
3. The difference (3 − 2i) − (−1 + 4i) is
- A4 − 6i
- B2 + 2i
- C4 + 2i
- D2 − 6i
Show answer
Answer: A. Subtract real parts and imaginary parts separately: 3 − (−1) = 4 and −2 − 4 = −6. Adding the second number instead gives 2 + 2i.
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