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The trigonometric form of a complex number

Lessons 39–40 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 10, Parts 1 and 2, 1st edition. EXTREMUM PRESS, Tashkent, 2017
39

Complex numbers of the form r(cos φ + i sin φ) and re^(iφ)

Textbook, Part 2: p. 80
GoalFind the modulus and argument of a complex number, write it in trigonometric and exponential form and convert back to algebraic form.
New words
modulus · modulargument · argumenttrigonometric form · trigonometrik shaklexponential form · ko‘rsatkichli shakl
Explanation

We show z = a + bi as the point (a, b) in the plane. The distance from the origin to this point, r = √(a² + b²), is the modulus |z|, and the angle φ from the positive Ox direction to this vector is the argument; a = r cos φ, b = r sin φ. The argument is taken in 0 ≤ φ < 2π and found from the quadrant of the point: do not rely on tan φ = b/a alone, since it gives wrong angles in quadrants II and III. z = r(cos φ + i sin φ) is the trigonometric form and z = r · e^(iφ) is the exponential form (Euler’s formula e^(iφ) = cos φ + i sin φ). For example i = cos π/2 + i sin π/2 = e^(iπ/2) and −1 = e^(iπ). To go from trigonometric to algebraic form, compute cos and sin and multiply by r. The argument of zero is undefined.

Worked examples
z = 1 + i: r = √(1 + 1) = √2; cos φ = 1/√2, sin φ = 1/√2, the point is in quadrant I, so φ = π/4. z = √2(cos π/4 + i sin π/4) = √2 · e^(iπ/4).
z = −√3 + i: r = √(3 + 1) = 2; cos φ = −√3/2, sin φ = 1/2, quadrant II, so φ = 5π/6. z = 2(cos 5π/6 + i sin 5π/6).
Class activity

On graph paper mark z = −√3 + i roughly (√3 ≈ 1.7), measure the angle of vector OA with a protractor and check that it is close to 150°, i.e. 5π/6 = 150°.

Practice
1
What is the formula for the modulus of z = a + bi?
2
Write z = 2(cos π/3 + i sin π/3) in algebraic form.
3
Find the modulus of z = 3 − 4i.
4
Why is the argument of z = −i equal to 3π/2?