|z| = √(1² + 1²) = √2. Argument = arctan(1/1) = π/4. Polar form: √2 ∠ 45°.

Find the Modulus and Argument of 1 + i — Complex Number Basics

Question

Find the modulus and argument of the complex number $z = 1 + i$ .

Express it in polar form as well.

Solution — Step by Step

Write $z = 1 + i$ in the standard form $a + bi$ .

Here, $a = 1$ (real part) and $b = 1$ (imaginary part).

The modulus $|z|$ is the distance from the origin to the point $(a, b)$ in the Argand plane.

|z| = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2}

The argument $\theta$ is the angle the line $OP$ makes with the positive real axis.

Since both $a > 0$ and $b > 0$ , the point $(1, 1)$ is in the first quadrant — the argument is simply $\arctan(b/a)$ :

\arg(z) = \arctan\left(\frac{1}{1}\right) = \arctan(1) = \frac{\pi}{4}

Polar form is $z = r(\cos\theta + i\sin\theta)$ , where $r = |z|$ and $\theta = \arg(z)$ .

z = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)

Final Answer: $|z| = \sqrt{2}$ , $\arg(z) = \dfrac{\pi}{4}$ (or 45°)

Why This Works

Every complex number $z = a + bi$ corresponds to a point $(a, b)$ on the Argand plane. The modulus is literally the length of that position vector — hence the Pythagoras formula.

The argument is the angle of that vector from the positive $x$ -axis. For points in the first quadrant, $\theta = \arctan(b/a)$ gives the correct angle directly. For other quadrants, we need to adjust — but $1 + i$ sits cleanly at 45° between the axes.

Polar form just packages this information neatly: instead of saying “1 unit right, 1 unit up”, we say “distance $\sqrt{2}$ , angle $\pi/4$ ”. Both describe the same point; polar form becomes essential when multiplying complex numbers.

Alternative Method — Euclid’s Shortcut for $\arctan(1)$

You don’t always need a calculator. We know that $\tan(45°) = 1$ , so whenever $a = b$ (both positive), the argument is immediately $\pi/4$ .

Memorise these “clean” arguments: if $b/a = 1$ , argument is $\pi/4$ . If $b/a = \sqrt{3}$ , argument is $\pi/3$ . If $b/a = 1/\sqrt{3}$ , argument is $\pi/6$ . These values appear repeatedly in NCERT exercises and JEE Main.

For $z = 1 + i$ : real and imaginary parts are equal, so without any calculation — $\arg(z) = \pi/4$ .

Common Mistake

Confusing quadrant adjustment. Students apply $\arctan(b/a)$ mechanically without checking the quadrant. For $z = -1 + i$ , the point is in the second quadrant, so $\arg(z) = \pi - \arctan(1) = 3\pi/4$ — NOT $\pi/4$ .

For $z = 1 + i$ , both parts are positive so no adjustment is needed. Always plot the point first; it takes two seconds and saves the entire mark.

A second trap: writing the argument as $45$ instead of $\frac{\pi}{4}$ . In CBSE board exams this is fine if you specify degrees. But JEE expects radians — so train yourself to write $\frac{\pi}{4}$ by default.

Question

Solution — Step by Step

Why This Works

Alternative Method — Euclid’s Shortcut for arctan⁡(1)\arctan(1)arctan(1)

Common Mistake

Try These Next

Alternative Method — Euclid’s Shortcut for $\arctan(1)$