The modulus of the following complex number is given by : \(Z=\dfrac{(1-2j)(3+j1)(1+j1)}{j(1-j1)}\)
\(\sqrt{50}\)
Never multiply the complex numbers out. The modulus is multiplicative, so it passes straight through products and quotients:
\(\left|\dfrac{z_1z_2z_3}{z_4z_5}\right|=\dfrac{|z_1||z_2||z_3|}{|z_4||z_5|}\)
Take the modulus of each factor separately and the whole problem becomes arithmetic with square roots.
Step 1 — each factor, using \(|a+jb|=\sqrt{a^{2}+b^{2}}\).
| Factor | Modulus |
|---|---|
| 1 − 2j | \(\sqrt{1+4}=\sqrt5\) |
| 3 + j1 | \(\sqrt{9+1}=\sqrt{10}\) |
| 1 + j1 | \(\sqrt{1+1}=\sqrt2\) |
| j | \(\sqrt{0+1}=1\) |
| 1 − j1 | \(\sqrt{1+1}=\sqrt2\) |
Step 2 — assemble.
\(|Z|=\dfrac{\sqrt5\cdot\sqrt{10}\cdot\sqrt2}{1\cdot\sqrt2}\)
The \(\sqrt2\) in the numerator cancels the one in the denominator:
\(|Z|=\sqrt5\cdot\sqrt{10}=\sqrt{50}\)
which is option 3, or 5√2 ≈ 7.07 if a decimal is wanted.
Two remarks worth carrying.
The pure imaginary j has modulus 1. Multiplying or dividing by j rotates a phasor by 90° but never changes its length — which is why the j in the denominator contributes nothing here. That single observation removes most of the arithmetic.
Option 4 is a trap for a different reading. A modulus can only be zero if one of the factors is zero, and none of these is. Whenever a "0" appears among the choices for a modulus, glance at the factors first; if all are non-zero, discard it immediately.
Why this matters beyond the arithmetic. Network and control transfer functions are exactly products of such factors, \(H(j\omega)=K\dfrac{(j\omega+z_1)(j\omega+z_2)}{(j\omega+p_1)(j\omega+p_2)}\). Taking the modulus factor by factor and the argument as a sum of angles,
\(\angle Z=\sum\angle(\text{numerator})-\sum\angle(\text{denominator})\)
is precisely what turns a Bode plot into a sum of straight-line segments.
Hence, \(|Z|=\sqrt{50}\).
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