IB Maths AI HLFurther Complex NumbersPaper 2 & 3AC circuits~9 min read
Frequency & Phase of Trig Functions
A sinusoidal function is the real or imaginary part of a complex exponential. That hidden connection turns the hard problem of adding two sine waves into a one-line job — perfect for AC voltages, where signals with the same frequency need to be combined.
📘 What you need to know
aeb(x−c)i = acos(b(x−c)) + aisin(b(x−c)).
acos(b(x−c)) = Re(aeb(x−c)i) and asin(b(x−c)) = Im(…).
AC voltages are sinusoidal: V = a sin(bt+c) or a cos(bt+c).
To add two sinusoidals: rewrite each as a complex exponential, factor the time-dependent part, combine the constant phasors, then take Re or Im at the end.
Use a GDC for the numerical phasor sum — results are usually irrational.
The method works only if the frequencies (the coefficient of x) match.
Sinusoidals as parts of complex exponentials
Every sinusoidal function hides inside a complex exponential. From the polar identity eiθ = cos θ + i sin θ, the cosine sits on the real axis and the sine sits on the imaginary axis. Multiply by an amplitude a and the same is true for aeiθ.
Sinusoidals from complex exponentialsaeb(x−c)i = acos(b(x−c)) + ai sin(b(x−c))
cos ↔ Re, sin ↔ Im
Examples: 3 cos(2x+5) = Re(3 e(2x+5)i) and 5 sin(3(x−2)) = Im(5 e3(x−2)i). The exponential carries the same amplitude and phase — only the wrapper changes.
Adding two sinusoidals of the same frequency
To add two cos or sin functions with a common frequency, swap each for its complex-exponential partner, factor out the common time-dependent piece ebxi, and use a GDC to combine the remaining constant phasors into a single AeiB. Re-attach the time factor and take Re or Im — the result is one sinusoidal of the same frequency, with a new amplitude A and new phase B.
Each sinusoidal is represented by a phasor — a vector whose length is the amplitude and whose angle is the phase. Adding the phasors as vectors gives the resultant amplitude and phase directly.
🧠Recipe — adding two sinusoidals of the same frequency
Check the frequencies match: the coefficient of x (or t) must be the same in both terms.
Rewrite each as a complex exponential: a cos(…) ↔ Re(ae(…)i); a sin(…) ↔ Im(ae(…)i).
Factor out ebxi, leaving a constant sum of phasors in the bracket.
Use the GDC to combine the bracket into one AeiB; re-attach the time factor.
Take Re for cos or Im for sin to land back in sinusoidal form: A cos(bx+B) or A sin(bx+B).
Worked examples
WE 1
Recognising sin and cos as parts of exponentials
Express each as the real or imaginary part of a complex exponential: (a) 5 sin(2x+1), (b) 7 cos(3x−2).
(a) sin ↔ Im5 sin(2x + 1) = Im(5 e(2x+1)i)(b) cos ↔ Re7 cos(3x − 2) = Re(7 e(3x−2)i)(a) Im(5 e(2x+1)i) · (b) Re(7 e(3x−2)i)amplitude in front, full argument inside the exponent.
WE 2
Adding two cosines with a π/2 phase gap
Write 4 cos(2t) + 6 cos(2t + π/2) in the form A cos(2t + B), giving exact A.
Write 3 sin(5t) + 4 sin(5t + π/3) in the form A sin(5t + B), giving exact A.
factor e5ti from the exponentialse5ti(3 + 4 eiπ/3)evaluate the phasor: 4 eiπ/3 = 2 + 2√3 i3 + 4 eiπ/3 = 5 + 2√3 imodulus & argument|.| = √(25 + 12) = √37; arg = tan⁻¹(2√3/5) ≈ 0.606take Im for sin√37 sin(5t + 0.606)A = √37 ≈ 6.08; B ≈ 0.606 rad.
WE 4
Combining two AC voltages
Two AC voltages are V1 = 12 sin(100t) and V2 = 16 sin(100t + π/2). Find the total voltage in the form V = A sin(100t + B).
factor e100tiz₁ + z₂ = e100ti(12 + 16 eiπ/2) = e100ti(12 + 16i)12 + 16i: a 12-16-20 right triangle|.| = √(144 + 256) = 20; arg = tan⁻¹(16/12) ≈ 0.927take Im for sinV = 20 sin(100t + 0.927)amplitude 20 V at phase 0.927 rad — cleanly a 12-16-20 triangle on the phasor diagram.
WE 5
Non-multiple-of-π phases (GDC needed)
Write 5 cos(2t + 1) + 3 cos(2t − 2) in the form A cos(2t + B), to 3 sf.
factor e2tie2ti(5 ei + 3 e−2i)evaluate the phasor on the GDC5 ei ≈ 2.702 + 4.207i3 e−2i ≈ −1.248 − 2.728isum ≈ 1.453 + 1.480imodulus & argument|.| ≈ 2.074; arg ≈ 0.7942.07 cos(2t + 0.794)irrational phases need the GDC — write each phasor in Cartesian first, then add.
WE 6
Full AC question with two non-zero phases
Two AC sources give V1 = 25 sin(50t + π/4) and V2 = 30 sin(50t + π/6). (a) Identify the related exponentials. (b) Find the total V = A sin(50t + B) to 3 sf.
(a) recognition: sin ↔ Imz₁ = 25 e(50t + π/4)i, z₂ = 30 e(50t + π/6)i(b) factor e50tie50ti(25 eiπ/4 + 30 eiπ/6)phasor on GDC25 eiπ/4 ≈ 17.68 + 17.68i30 eiπ/6 ≈ 25.98 + 15.00isum ≈ 43.66 + 32.68imodulus & argument|.| ≈ 54.5; arg ≈ 0.642V ≈ 54.5 sin(50t + 0.642)two non-trivial phases — the workflow is identical, the arithmetic stays on the GDC.
💡 Top tips
Same frequency required: 2 sin(3x+1) + 3 sin(3x−5) is fine; 2 sin(3x+1) + 3 sin(5x+1) is not.
Stay consistent: if the originals are both sines, take Im at the end; both cosines ⇒ take Re.
Factor the time piece first — the remaining bracket is constant and goes straight on the GDC.
For exact phases like π/3, π/4, π/6 the phasor has exact Cartesian parts — use them.
Use your GDC’s polar / rectangular conversions to add phasors and read off modulus and argument.
âš Common mistakes
Adding two sinusoidals with different frequencies — this method fails unless the coefficients of x agree.
Taking Re of a sin problem (or Im of a cos problem) — sin ↔ Im, cos ↔ Re; mismatch gives the wrong answer.
Forgetting to factor ebxi first — without it the GDC can’t handle the symbolic t.
Adding phases directly — the new phase comes from the resultant phasor’s argument, not from averaging the originals.
Using degrees — complex-exponential phases are in radians.
That completes Further Complex Numbers. From a single insight — sinusoidals are real or imaginary parts of complex exponentials — the whole machinery of phasors and AC analysis falls out, with the GDC doing the arithmetic and the exponential rules doing the structure.
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