For z₁ = 6 − j3 and z₂ = 1 + j2, find the real part of w = 3z₁ + 2z̄₂.
Complex numbers worksheet with answers
1.Easy
z₁ = 6 − j3z₂ = 1 + j2
Re(w) =
2.Easy
For z₁ = −5 + j and z₂ = −4 − j3, find the modulus of the product, |z₁z₂|.
z₁ = −5 + jz₂ = −4 − j3
|z₁z₂| =
3.Medium
The diagram shows z = −1 − j2. With w = 2∠90°, click the point that represents z·w.
z = −1 − j2w = 2∠90°
- 1A (2 − j)
- 2B (−4 + j2)
- 3C (4 − j2)
- 4D (−2 − j4)
4.Medium
Find the argument θ of z = −7 + j8 in degrees, as the principal value between −180° and 180°.
z = −7 + j8
θ = °
5.Hard
One root of z³ − z² − 15z − 25 = 0 is z₁ = −2 + j. The coefficients are real. Find the real root r.
z₁ = −2 + j
r =
6.Hard
One root of z³ − 6z² + 28z − 40 = 0 is z₁ = 2 + j4. The coefficients are real. Find the real root r.
z₁ = 2 + j4
r =
Answer key
- 1. Re(w) = 20
- 2. |z₁z₂| = 25.5
- 3. C (4 − j2)
- 4. θ = 131.2 °
- 5. r = 5
- 6. r = 2
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