Phase-Shifting Network with a Real Dial
An electronics laboratory is tuning a phase-shifting network whose normalized impedance is z = (k + 3i)/(1 − ki), where k is a real-number setting on a dial and i² = −1. A technician claims that |z| must be constant because the k-dependent terms in the numerator and denominator cancel. A second technician claims that no real setting of k can make z purely imaginary. The supervisor asks for a written check before the network is connected to a sensitive amplifier. The dial is calibrated for all real k, including k = 0, and the team must know whether any real k makes z real, whether any makes z purely imaginary, how |z| behaves as k is varied, and what curve the point representing z traces in the Argand plane. The final note must also state any point on that curve that the circuit cannot actually produce with a finite dial setting.
1. Write z in the form x + yi and decide which of the two technician claims is correct. In particular, state whether any real k makes z real or purely imaginary.
Read the worked answer
Multiply numerator and denominator by the conjugate 1 + ki:
z = (k + 3i)(1 + ki)/(1 + k²) = (k + k²i + 3i + 3ki²)/(1 + k²) = (k − 3k + i(k² + 3))/(1 + k²) = −2k/(1 + k²) + i(k² + 3)/(1 + k²).
Thus x = −2k/(1 + k²) and y = (k² + 3)/(1 + k²). For z to be real, y = 0, which needs k² + 3 = 0: impossible for real k. For z to be purely imaginary, x = 0, so k = 0; then z = 3i. So the first claim about constant |z| is false, and the second claim is also false-k = 0 gives a purely imaginary value. No real k makes z real.
2. Show that |z|² = (k² + 9)/(k² + 1), find the range of |z| for real k, and state whether the value 1 is attained.
Read the worked answer
From the real and imaginary parts, |z|² = [4k² + (k² + 3)²]/(1 + k²)² = (k⁴ + 10k² + 9)/(1 + k²)² = [(k² + 1)(k² + 9)]/(k² + 1)² = (k² + 9)/(k² + 1).
Write this as 1 + 8/(k² + 1). Since k² ≥ 0, the denominator is at least 1, so 8/(k² + 1) is positive and decreases as |k| increases. Its maximum is 8 at k = 0, giving |z|² = 9, i.e. |z| = 3. As |k| → ∞, |z|² → 1, but for every finite k the value is greater than 1. Hence 1 < |z| ≤ 3. The value |z| = 1 is not attained by any finite real k.
3. Describe the locus of the point representing z in the Argand plane, and identify any point of the locus not produced by a finite k.
Read the worked answer
Let z = x + yi. From above, y − 1 = [(k² + 3) − (k² + 1)]/(k² + 1) = 2/(k² + 1), and x² = 4k²/(1 + k²)². Then x² + (y − 1)² = 4k²/(1 + k²)² + 4/(1 + k²)² = 4(k² + 1)/(1 + k²)² = 4/(1 + k²) = 2(y − 1). So x² + y² − 2y + 1 = 2y − 2, hence x² + y² − 4y + 3 = 0, i.e. x² + (y − 2)² = 1. The locus is the circle with centre (0, 2) and radius 1. Since y = 1 + 2/(1 + k²) > 1 for every finite k, the lowest point (0, 1) of the circle is not produced; it is approached as |k| → ∞.