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Question

The modulus of the following complex number is given by :

\(Z=\dfrac{(1-2j)(3+j1)(1+j1)}{j(1-j1)}\)

This question was previously asked in
UGC NET 2015 Paper 1 Question Paper (27-Dec-2015)
The correct answer is

\(\sqrt{50}\)

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 \(|a+jb|=\sqrt{a^{2}+b^{2}}\).

FactorModulus
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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