For z₁ = 2 − j3 and z₂ = 5 + j, find the real part of w = −z₁ − 2z̄₂.
Complex numbers worksheet with answers
1.Easy
z₁ = 2 − j3z₂ = 5 + j
Re(w) =
2.Easy
For z₁ = −7 + j and z₂ = 1 − j3, find the imaginary part of the product z₁·z₂.
z₁ = −7 + jz₂ = 1 − j3
Im(z₁z₂) =
3.Medium
Write z = 2.5∠−40° in rectangular form a + jb and give the real part a.
r = 2.5θ = −40°
a =
4.Medium
The diagram shows z = 1 − j. With w = 0.5∠90°, click the point that represents z/w.
z = 1 − jw = 0.5∠90°
- 1A (2 − j2)
- 2B (2 + j2)
- 3C (−2 − j2)
- 4D (−1 − j)
5.Hard
Exactly one of the points A–D is a root of z³ = 8∠60°. Click it.
w = 8∠60°n = 3
- 1A (1∠−100°)
- 2B (2∠100°)
- 3C (2∠−100°)
- 4D (2∠−130°)
6.Hard
One root of z³ − 2z² − 6z + 72 = 0 is z₁ = 3 + j3. The coefficients are real. Find the real root r.
z₁ = 3 + j3
r =
Answer key
- 1. Re(w) = −12
- 2. Im(z₁z₂) = 22
- 3. a = 1.915
- 4. C (−2 − j2)
- 5. C (2∠−100°)
- 6. r = −4
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