If z = (1+i)/(1-i), find |z| and arg(z)

Question

If $z = \dfrac{1+i}{1-i}$ , find $|z|$ and $\arg(z)$ .

(JEE Main 2022, similar pattern)

Solution — Step by Step

Multiply numerator and denominator by the conjugate of the denominator, which is $(1+i)$ :

z = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)}

Denominator: $(1-i)(1+i) = 1 - i^2 = 1 - (-1) = 2$

Numerator: $(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$

z = \frac{2i}{2} = i

$z = i = 0 + 1 \cdot i$

This is the point $(0, 1)$ on the Argand plane — sitting on the positive imaginary axis.

|z| = \sqrt{0^2 + 1^2} = \mathbf{1}

\arg(z) = \mathbf{\frac{\pi}{2}}

Why This Works

Rationalising a complex fraction means eliminating $i$ from the denominator. Multiplying by the conjugate works because $(a + bi)(a - bi) = a^2 + b^2$ , which is always real and positive. Once the denominator is real, we can separate the result into $x + iy$ form and directly read the modulus and argument.

The result $z = i$ makes geometric sense too. Dividing $(1+i)$ by $(1-i)$ is the same as multiplying by $\frac{1}{1-i}$ , which has argument $\pi/4$ . Since $(1+i)$ itself has argument $\pi/4$ , the total argument is $\pi/4 + \pi/4 = \pi/2$ . Both numbers have modulus $\sqrt{2}$ , so the modulus of the quotient is $\sqrt{2}/\sqrt{2} = 1$ .

Alternative Method — Use modulus and argument properties directly

Without simplifying $z$ :

|z| = \frac{|1+i|}{|1-i|} = \frac{\sqrt{2}}{\sqrt{2}} = 1

\arg(z) = \arg(1+i) - \arg(1-i) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}

This method is faster and avoids the algebra entirely.

For JEE MCQs, always try the modulus/argument shortcut first. If the question only asks for $|z|$ or $\arg(z)$ , you can get the answer in 15 seconds without rationalising. Save the full simplification for when you need the Cartesian form $x + iy$ .

Common Mistake

Students sometimes write $\arg(1-i) = \pi/4$ instead of $-\pi/4$ . The number $1-i$ lies in the fourth quadrant (positive real, negative imaginary), so its argument is $-\pi/4$ (or equivalently $7\pi/4$ ). Mixing up the quadrant of $1-i$ with $1+i$ flips the final answer. Always plot the number mentally on the Argand plane before assigning the argument.

Question

Solution — Step by Step

Why This Works

Alternative Method — Use modulus and argument properties directly

Common Mistake

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