Complex numbers: practice problems with step-by-step solutions
Also called: imaginary numbers · Argand diagram · De Moivre’s theorem · polar form
Free Complex numbers practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: arithmetic in rectangular form, polar and exponential form, De Moivre, roots and equations.
Free practice problemEasy
For z₁ = 1 + j4 and z₂ = 5 + j6, find the imaginary part of w = 2z₁ + 5z̄₂.
The equation x² = −1 has no real solution. Engineers add a new number j with j² = −1. Mathematicians write i; engineering uses j because i already means current.
A complex number has the form z = a + jb. The number a is the real part and b is the imaginary part: Re(z) = a and Im(z) = b.
You can draw z as the point (a, b) in the Argand diagram: the real part along the horizontal axis, the imaginary part along the vertical one.
The distance from the origin is the modulus |z| = √(a² + b²). The angle from the positive real axis is the argument θ. The mirror image in the real axis, z̄ = a − jb, is the conjugate.
Self-check What is the modulus of z = 3 − j4?
25
5
7
−1
Answer: 5. |z| = √(3² + (−4)²) = √25 = 5. Adding 3 and 4 gives 7, and forgetting the square root gives 25.
2Adding, multiplying and dividing
Add and subtract part by part: (2 + j3) + (1 − j4) = 3 − j.
Answer: 2. It is a number times its conjugate: 1² + 1² = 2. Writing 1 − j² as 1 − 1 = 0 forgets that j² = −1, so −j² = +1.
3Polar and exponential form
A point can also be given by its distance and its angle: z = r∠θ, with r = |z| and θ = arg z. Back to rectangular form with a = r cos θ and b = r sin θ.
z = r∠θ = r(cos θ + j sin θ) = r·ejθ
The last form uses Euler’s formula ejθ = cos θ + j sin θ. In the exponential form θ must be in radians.
Watch the quadrant. A calculator gives atan(b/a) between −90° and 90°. If a < 0 the point is on the left, so add or subtract 180°.
Example: z = −2 + j2. Then r = √8 = 2.828. The reference angle is atan(2/2) = 45°, and the point is in the second quadrant, so θ = 180° − 45° = 135°.
Self-check What is the argument of z = −1 − j, between −180° and 180°?
135°
45°
−135°
−45°
Answer: −135°. The point lies in the third quadrant. The reference angle is 45°, so θ = −(180° − 45°) = −135°. atan(−1/−1) = 45° points to the opposite quadrant.
4Multiplying in polar form, powers and roots
In polar form multiplication is easy: multiply the moduli and add the arguments. For division, divide the moduli and subtract the arguments.
(r₁∠θ₁)(r₂∠θ₂) = r₁r₂∠(θ₁ + θ₂)
Repeating this gives De Moivre’s theorem: (r∠θ)ⁿ = rⁿ∠nθ.
The equation zⁿ = R∠φ has exactly n roots. They all have the modulus R1/n and the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1.
For z³ = 8 = 8∠0°: the modulus is 2 and the arguments are 0°, 120° and 240° (that is −120°). The roots form a regular triangle on the circle.
A quadratic z² + pz + q = 0 with p² < 4q has the complex roots −p/2 ± j√(q − p²/4). Real polynomials always have their complex roots in conjugate pairs.
5A worked example
Powers of numbers like 1 + j are much faster in polar form than by multiplying out.
Worked example
Compute (1 + j)6 with De Moivre’s theorem.
1
Write 1 + j in polar form
r = √(1² + 1²) = √2. The point is in the first quadrant, so θ = atan(1/1) = 45°.
1 + j = √2∠45°
2
Apply De Moivre
Raise the modulus to the power 6 and multiply the angle by 6.
(√2)6 = 8, 6·45° = 270°
3
Bring the angle into range
270° is the same direction as −90°.
(1 + j)6 = 8∠−90°
4
Back to rectangular form
8(cos(−90°) + j sin(−90°)) = 8(0 − j) = −j8
Answer
(1 + j)6 = −j8: real part 0, imaginary part −8.
6Common mistakes and exam questions
Forgetting that j² = −1 when you multiply out, so a term ends up with the wrong sign.
Dividing part by part: (a + jb)/(c + jd) is not a/c + jb/d. Use the conjugate.
Taking atan(b/a) as the argument when a < 0. Sketch the point first.
Using degrees in ejθ, or leaving the calculator in radians for r∠θ given in degrees.
In De Moivre: forgetting to raise the modulus to the power, or giving only one root of zⁿ = w.
Easy questions ask for a part of a product, quotient or sum. Medium ones convert between rectangular, polar and exponential form. Hard ones use De Moivre, the n-th roots and polynomial equations.
Self-check z₁ = 2∠30° and z₂ = 4∠50°. What is z₁·z₂?
6∠80°
8∠1500°
8∠20°
8∠80°
Answer: 8∠80°. Multiply the moduli (2·4 = 8) and add the arguments (30° + 50° = 80°). Adding the moduli gives 6, and multiplying the angles makes no sense.
For z₁ = −7 + j and z₂ = 4 + j2, find the real part of the product z₁·z₂.
z₁ = −7 + jz₂ = 4 + j2
Find Re(z₁z₂)
Show hints
Multiply out the brackets like (a + b)(c + d), giving four terms.
Replace j² by −1 so that one of the four terms becomes real.
(−7 + j)(4 + j2) = (−28) + j(−14) + j4 + j²·2
Show the answer and the step-by-step solution
Complex numbers: z = a + jb has real part a and imaginary part b, with j² = −1 (mathematicians write i). Add and subtract part by part; multiply out the brackets and replace j² by −1; divide by multiplying the top and bottom by the conjugate of the denominator, z̄ = a − jb, because z·z̄ = a² + b² is real. On the Argand diagram z is the point (a, b): its distance from the origin is the modulus |z| = √(a² + b²), and the angle from the positive real axis is the argument θ.
Write z = 4∠160° in rectangular form a + jb and give the imaginary part b.
r = 4θ = 160°
Find b
Show hints
a = r·cos θ and b = r·sin θ.
sin(160°) = 0.342
Multiply by r = 4.
Show the answer and the step-by-step solution
Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.
Components
a = 4·cos(160°) = −3.759
b = 4·sin(160°) = 1.368
Rectangular form
z = −3.759 + j1.368
b = 1.368
Answer
b = 1.368
Example 3Hard
The roots of zⁿ = w with n = 4 and w = 16∠−60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the real part of z₂.
w = 16∠−60°n = 4k = 2
Find Re(z₂)
Show hints
Every root has the modulus R1/n = 161/4 = 2.
The argument of z₂ is (φ + 360°·k)/n with φ = −60°, k = 2 and n = 4.
a = R1/n·cos(165°).
Show the answer and the step-by-step solution
De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.
Modulus of the roots
161/4 = 2
Argument of the root
θ₂ = (−60° + 360°·2)/4 = 165°
Rectangular form
z₂ = 2·(cos 165° + j sin 165°) = −1.932 + j0.5176
Re(z₂) = −1.932
Answer
Re(z₂) = −1.932
The 4 roots lie on the circle of radius 2, 90° apart.
Questions and answers
What is Complex numbers?
The equation x² = −1 has no real solution. Engineers add a new number j with j² = −1. Mathematicians write i; engineering uses j because i already means current.
A complex number has the form z = a + jb. The number a is the real part and b is the imaginary part: Re(z) = a and Im(z) = b.
How do I solve Complex numbers problems step by step?
Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).
What are common mistakes in Complex numbers?
Treating j² as +1 when multiplying out.
Dividing the real and imaginary parts separately.
Taking atan(b/a) as the argument for a point in the left half-plane.
Mixing degrees and radians, especially in ejθ.
Giving only one root of zⁿ = w, or forgetting to take the n-th root of the modulus.
What do the Complex numbers exercises cover?
Easy: arithmetic in rectangular form
Medium: polar and exponential form
Hard: De Moivre, roots and equations
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.