Convert between Cartesian and polar complex forms
Use modulus and argument to express a non-zero number as r(cos θ + i sin θ) or re^(iθ), and recover its Cartesian components.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.9. Paper 1 algebra and radian trigonometry are assumed. Use Cartesian working for products, quotients and the two square roots of a non-zero complex number. Arguments usually use −π < θ ≤ π; an alternative interval 0 ≤ θ < 2π is acceptable unless specified. The argument of zero is undefined. General nth-root methods are beyond this mapped sequence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Find modulus and argument on an Argand diagram.
- Use exact sine and cosine values in radians.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which expression means cis θ?
What you will learn
- Convert Cartesian components to modulus-argument form with a valid quadrant.
- Recover exact Cartesian components from trigonometric or exponential form.
Polar form stores a length and a direction
For non-zero z = x + iy, set r = |z| > 0 and choose an argument θ. Then x = rcos θ and y = rsin θ, giving z = r(cos θ + i sin θ). Exponential notation re^(iθ) represents the same complex value. A useful abbreviation is cis θ = cos θ + i sin θ, but write out its meaning when first used. The number 1 + i√3 is 2cis(π/3), since the coordinates are 2·1/2 and 2·√3/2.
Use the quadrant and the stated interval when choosing θ; arctan(y/x) alone cannot distinguish opposite points. For x = 0 choose the positive or negative imaginary direction directly instead of dividing by zero. The modulus is non-negative; using a negative radius requires changing the direction and is not the standard modulus-argument representation.
Convert back using both trigonometric components
For 3cis(7π/4), cos(7π/4) = √2/2 and sin(7π/4) = −√2/2, so Cartesian form is 3√2/2 − (3√2/2)i. Its principal argument is −π/4, equivalent modulo 2π. Replacing 7π/4 by π/4 would change the imaginary sign and the point. An exact radian angle is preferable to a rounded decimal during subsequent operations.
Although zero can be written with r = 0 and any trigonometric angle, this does not give it an argument: every such expression equals the origin. The non-zero restriction is needed for directional statements and polar division. Do not confuse e^(iθ) with e^θ; the imaginary exponent encodes the cosine and sine pair, rather than ordinary real exponential growth.
Pause and explain
Which polar form represents a point in quadrant III?
Worked example
Express 1 + i√3 in trigonometric and exponential form.
Show the worked solution
- Its modulus is √(1 + 3) = 2.
- Quadrant I gives θ = π/3.
- Use x = rcos θ and y = rsin θ to verify both components.
Answer 2(cos(π/3) + i sin(π/3)) = 2e^(iπ/3).
Try a topic test
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Express √3 + i in modulus-argument and exponential form using the principal argument.
Give me a hint
Calculate the length and use the quadrant I exact angle with tangent 1/√3.
Compare my reasoning
- The modulus is √(3 + 1) = 2.
- The principal argument is π/6.
- The components of 2cis(π/6) are √3 and 1, so the exponential form is 2e^(iπ/6).
2cis(π/6) = 2e^(iπ/6).
Look for these in your work
- Use the exact modulus and radian angle.
- Check both Cartesian components.
Convert −2 − 2√3i to polar form with a principal argument, and give an equivalent angle in 0 ≤ θ < 2π.
Give me a hint
Its reference angle is π/3 but both coordinates are negative.
Compare my reasoning
- The modulus is √(4 + 12) = 4.
- Quadrant III gives principal argument −2π/3.
- Adding 2π gives 4π/3; both have cosine −1/2 and sine −√3/2.
4cis(−2π/3); equivalent angle 4π/3.
Look for these in your work
- Check both signs when selecting the quadrant.
- Give the equivalent angle in the requested interval.
Convert 3e^(7πi/4) to exact Cartesian form and state its principal argument.
Give me a hint
Use the signs of cosine and sine at the given quadrant IV angle.
Compare my reasoning
- Write 3[cos(7π/4) + i sin(7π/4)].
- Use cosine √2/2 and sine −√2/2 to obtain 3√2/2 − (3√2/2)i.
- Subtracting 2π from 7π/4 gives principal argument −π/4.
3√2/2 − (3√2/2)i; principal argument −π/4.
Look for these in your work
- Preserve the exact components and imaginary sign.
- Normalise the supplied angle without changing its direction.
Common mistakes
- Use a non-negative modulus and a direction in the correct quadrant.
- Preserve the sign of sine when converting back.
- A zero modulus does not define an argument for zero.
What can you explain now?
What is the Cartesian form of 2e^(iπ/2)?
Compare with the explanation
2i.
The cosine component is zero and the sine component is 2.
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Make it stick
Convert a non-zero complex value between Cartesian, trigonometric and exponential forms, checking radius, angle interval and both coordinate signs.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.