MHT-CET Maths · Complex Numbers

Modulus and Argument — Polar Form, De Moivre and Square Roots

|z| is the distance from the origin and arg z the angle from the positive real axis — the modulus multiplies and divides, the argument adds and subtracts, and z = r(cos θ + i sin θ) makes powers routine.

All Complex Numbers notes MHT-CET Maths strategy

Why this matters

26 PYQs at 19% HARD — one of the chapter's two softest pages, which is why it is the half worth owning. The modulus of a product or quotient of factors is asked every year and needs no expansion at all; the argument questions are wrong only when the quadrant is ignored; and |z| + z = a + ib has been set four times with two different right-hand sides. Three stems here were repaired against the papers this session: a magnitude that belonged to the twin sitting, an argument denominator, and a key that pointed at the twin's answer.

Concept 1 of 5: Modulus and Its Properties: |z₁z₂| = |z₁||z₂|

|z| = \sqrt{x^2 + y^2}

is the length of the arrow from

0

z

. Lengths multiply when complex numbers multiply and divide when they divide — so the modulus of a big product is computed factor by factor, never by expanding.

Definition

$|x + iy| = \sqrt{x^2 + y^2}$ ; $|\bar z| = |z|$ ; $|z|^2 = z\bar z$ .
$|z_1z_2| = |z_1||z_2|$ , $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$ , $|z^n| = |z|^n$ . So $\left|\dfrac{(\sqrt3 + i)^3(4 + 3i)^2}{(8 + 6i)^2}\right| = \dfrac{2^3\cdot5^2}{10^2} = 2$ .
$|z_1 + z_2| \ne |z_1| + |z_2|$ in general (triangle inequality: $\le$ ).
Modulus of a square root: if $w^2 = z$ then $|w|^2 = |z|$ , so $|\sqrt z| = \sqrt{|z|}$ . For $6 + 8i$ : $\sqrt{10}$ ; for the conjugate of $-7 + 24i$ : $\sqrt{25} = 5$ .
Memorise the Pythagorean moduli: $|3 + 4i| = 5$ , $|8 + 6i| = 10$ , $|5 + 12i| = 13$ , $|7 + 24i| = 25$ , $|\sqrt3 + i| = 2$ , $|1 + i| = \sqrt2$ .

Modulus rules

|z_1z_2| = |z_1||z_2| \qquad \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \qquad |z^n| = |z|^n \qquad |\sqrt z| = \sqrt{|z|}

Worked example

Find

|z|

for

z = \dfrac{(1 + i)^4(5 + 12i)}{(3 - 4i)^2}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q144Moderate

Example 1 · Complex Numbers · Modulus and Argument — Polar Form, De Moivre and Square Roots

z = \dfrac{(\sqrt{3}+i)^3(3i+4)^2}{(8+6i)^2}

, then

|z|

is equal to

Expanding the product to find its modulus

(\sqrt3 + i)^3(4 + 3i)^2

expanded is a page of algebra with several sign traps. Moduli multiply:

2^3\cdot5^2

. Never expand for a modulus.

Drill 2 more on modulus and its properties: |z₁z₂| = |z₁||z₂|

Concept 2 of 5: The Argument: Reference Angle Plus the Quadrant

\tan^{-1}\dfrac{y}{x}

only gives a reference angle; the quadrant of the point decides whether to keep it, add

\pi

, or subtract from

\pi

-\dfrac12 + \dfrac{\sqrt3}{2}i

is in the second quadrant, so its argument is

\pi - \dfrac{\pi}{3}

, not

-\dfrac{\pi}{3}

Definition

$\arg z = \theta$ with $\cos\theta = \dfrac{x}{|z|}$ , $\sin\theta = \dfrac{y}{|z|}$ ; principal value in $(-\pi, \pi]$ .
Quadrant rule with reference angle $\alpha = \tan^{-1}\left|\dfrac{y}{x}\right|$ : I → $\alpha$ ; II → $\pi - \alpha$ ; III → $-(\pi - \alpha)$ (or $\pi + \alpha$ if $[0, 2\pi)$ is used); IV → $-\alpha$ .
Rationalise first when $z$ is a fraction: $\dfrac{-2}{1 + \sqrt3 i} = -\dfrac12 + \dfrac{\sqrt3}{2}i$ , argument $\dfrac{2\pi}{3}$ .
$\arg(z_1z_2) = \arg z_1 + \arg z_2$ , $\arg\dfrac{z_1}{z_2} = \arg z_1 - \arg z_2$ : $\arg\dfrac{1 + i\sqrt3}{\sqrt3 - i} = \dfrac{\pi}{3} - \left(-\dfrac{\pi}{6}\right) = \dfrac{\pi}{2}$ — or rationalise to $i$ directly.
When the argument is not a standard angle, leave it as $\tan^{-1}$ of the simplified ratio: $\dfrac{z_1 + z_2}{z_1 - z_2} = \dfrac{19 + 22i}{13}$ has argument $\tan^{-1}\dfrac{22}{19}$ .

Argument rules

\arg(z_1z_2) = \arg z_1 + \arg z_2 \qquad \arg\frac{z_1}{z_2} = \arg z_1 - \arg z_2 \qquad \arg\bar z = -\arg z

Worked example

Find the argument of

z = \dfrac{-4}{1 + i}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q140Moderate

Example 2 · Complex Numbers · Modulus and Argument — Polar Form, De Moivre and Square Roots

Z=\frac{-2}{1+\sqrt{3}i}

i=\sqrt{-1}

, then the value of

\arg(Z)

The argument from the ratio alone

\tan^{-1}\dfrac{\sqrt3/2}{-1/2} = \tan^{-1}(-\sqrt3) = -\dfrac{\pi}{3}

is the fourth-quadrant answer for a second-quadrant point. Plot the point first; the un-adjusted angle is always an option.

Drill 2 more on the argument: reference angle plus the quadrant

Concept 3 of 5: Polar Form and De Moivre: Powers, Rotations and sin θ + i cos θ

Write

z = r(\cos\theta + i\sin\theta)

. Then

z^n = r^n(\cos n\theta + i\sin n\theta)

: multiply the angle, raise the modulus. Multiplying by

i

is a rotation by

\dfrac{\pi}{2}

, which is how a 'particle moves through an angle' stem is answered.

Definition

Polar coordinates: $(-2\sqrt3, 2)$ has $r = 4$ and, being in quadrant II, $\theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ .
De Moivre: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for any integer $n$ .
$\sin\theta + i\cos\theta = i(\cos\theta - i\sin\theta) = i\,e^{-i\theta}$ — NOT a polar form as written. So $\dfrac{(\cos\theta + i\sin\theta)^4}{(\sin\theta + i\cos\theta)^5} = \dfrac{e^{4i\theta}}{i\,e^{-5i\theta}} = -i\,e^{9i\theta} = \sin 9\theta - i\cos 9\theta$ .
Rotation: multiplying $z$ by $i$ rotates it $\dfrac{\pi}{2}$ anticlockwise about the origin: $(x, y) \to (-y, x)$ . Translations are additions: 'moves $5$ units horizontally' is $z + 5$ ; ' $\sqrt2$ units along $\hat i + \hat j$ ' is $z + (1 + i)$ .

De Moivre and the rotation by i

(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \qquad iz \text{ is } z \text{ rotated by } \tfrac{\pi}{2} \qquad \sin\theta + i\cos\theta = i(\cos\theta - i\sin\theta)

Worked example

Evaluate

\left(\cos\dfrac{\pi}{9} + i\sin\dfrac{\pi}{9}\right)^{6}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift II · Q140Hard

Example 3 · Complex Numbers · Modulus and Argument — Polar Form, De Moivre and Square Roots

The value of

\frac{(\cos\theta+isin\theta)^{4}}{(\sin\theta+icos\theta)^{5}}=

where

i =\sqrt{- 1}

Applying De Moivre to sin θ + i cos θ

(\sin\theta + i\cos\theta)^5

is NOT

\sin 5\theta + i\cos 5\theta

. Rewrite it as

i(\cos\theta - i\sin\theta)

first; the

i^5 = i

in front is where the final

\sin 9\theta - i\cos 9\theta

comes from.

Drill 2 more on polar form and de moivre: powers, rotations and sin θ + i cos θ

Concept 4 of 5: |z| + z = a + ib: Equate the Imaginary Part, Then Solve for |z|

|z|

is real, so in

|z| + z = a + ib

the imaginary part of

z

must be

b

, and

|z| + x = a

. Substituting

|z| = \sqrt{x^2 + b^2}

gives one equation in

x

whose

x^2

terms cancel.

Definition

Put $z = x + iy$ : $y = b$ and $\sqrt{x^2 + b^2} = a - x$ .
Square: $x^2 + b^2 = a^2 - 2ax + x^2 \Rightarrow x = \dfrac{a^2 - b^2}{2a}$ , and then $|z| = a - x = \dfrac{a^2 + b^2}{2a}$ .
$|z| + z = 2 + i$ : $|z| = \dfrac{4 + 1}{4} = \dfrac54$ . $|z| + z = 3 + i$ : $|z| = \dfrac{9 + 1}{6} = \dfrac53$ .
Both versions have been set; the two answers $\dfrac54$ and $\dfrac53$ appear in each other's option lists, and the 2022 sitting's stored key once pointed at the wrong one.

Closed form

|z| + z = a + ib \ \Rightarrow\ \operatorname{Im}z = b,\quad |z| = \frac{a^2 + b^2}{2a}

Worked example

|z| + z = 4 + 2i

, find

|z|

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q110Moderate

Example 4 · Complex Numbers · Modulus and Argument — Polar Form, De Moivre and Square Roots

Let

z

be a complex number such that

|z|+z = 2+i

, where

i=\sqrt{-1}

, then

|z|

is equal to

Answering the twin sitting's value

\frac54

belongs to

2 + i

and

\frac53

3 + i

; each list offers both. Read the right-hand side before recalling the number.

Drill 3 more on |z| + z = a + ib: equate the imaginary part, then solve for |z|

Concept 5 of 5: Find z From a Given Modulus: Simplify, Then Fix the Parameter

z = \dfrac{(1 + i)^2}{a - i}

simplifies to

\dfrac{2i}{a - i}

, whose modulus is

\dfrac{2}{\sqrt{a^2 + 1}}

. A given magnitude fixes

a

; then rationalising gives

z

, and the stem may ask for

z

\bar z

Definition

Simplify the numerator: $(1 + i)^2 = 2i$ . Then $|z| = \dfrac{|2i|}{|a - i|} = \dfrac{2}{\sqrt{a^2 + 1}}$ — moduli divide, no rationalising needed yet.
$|z| = \dfrac{2}{\sqrt5}$ gives $a = 2$ ; $|z| = \sqrt{\dfrac25}$ gives $a = 3$ . Two sittings used the two magnitudes, and their option lists differ accordingly.
Then $z = \dfrac{2i(a + i)}{a^2 + 1} = \dfrac{-2 + 2ai}{a^2 + 1}$ : with $a = 2$ , $z = -\dfrac25 + \dfrac45 i$ ; with $a = 3$ , $z = -\dfrac15 + \dfrac35 i$ and $\bar z = -\dfrac15 - \dfrac35 i$ .
A modulus expression can also replace the algebra entirely: for $a + ib = \dfrac{3}{2 + \cos\theta + i\sin\theta}$ , $(a - 2)^2 + b^2 = |a + ib - 2|^2 = \left|\dfrac{3 - 2w}{w}\right|^2$ with $w = 2 + e^{i\theta}$ ; since $|3 - 2w| = |{-1} - 2e^{i\theta}| = |w|$ (both equal $\sqrt{5 + 4\cos\theta}$ ), the value is $1$ .

Modulus first, algebra second

\left|\frac{(1 + i)^2}{a - i}\right| = \frac{2}{\sqrt{a^2 + 1}} \qquad |2 + \cos\theta + i\sin\theta|^2 = 5 + 4\cos\theta

Worked example

a > 0

and

z = \dfrac{(1 - i)^2}{a + i}

has modulus

\dfrac{2}{\sqrt{10}}

, find

z

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q136Moderate

Example 5 · Complex Numbers · Modulus and Argument — Polar Form, De Moivre and Square Roots

a>0

and

z=\dfrac{(1+i)^2}{a-i}

i=\sqrt{-1}

, has magnitude

\dfrac{2}{\sqrt{5}}

, then

z

z or z̄?

The 2023 sitting asks for

\bar z

and the 2024 sitting for

z

, with sign-flipped options in each. Compute

z

, then read the last word of the stem before choosing.

Drill 3 more on find z from a given modulus: simplify, then fix the parameter

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Modulus and Its Properties: |z₁z₂| = |z₁||z₂|
Modulus rules
$|z_1z_2| = |z_1||z_2| \qquad \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \qquad |z^n| = |z|^n \qquad |\sqrt z| = \sqrt{|z|}$
The Argument: Reference Angle Plus the Quadrant
Argument rules
$\arg(z_1z_2) = \arg z_1 + \arg z_2 \qquad \arg\frac{z_1}{z_2} = \arg z_1 - \arg z_2 \qquad \arg\bar z = -\arg z$
Polar Form and De Moivre: Powers, Rotations and sin θ + i cos θ
De Moivre and the rotation by i
$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \qquad iz \text{ is } z \text{ rotated by } \tfrac{\pi}{2} \qquad \sin\theta + i\cos\theta = i(\cos\theta - i\sin\theta)$
|z| + z = a + ib: Equate the Imaginary Part, Then Solve for |z|
Closed form
$|z| + z = a + ib \ \Rightarrow\ \operatorname{Im}z = b,\quad |z| = \frac{a^2 + b^2}{2a}$
Find z From a Given Modulus: Simplify, Then Fix the Parameter
Modulus first, algebra second
$\left|\frac{(1 + i)^2}{a - i}\right| = \frac{2}{\sqrt{a^2 + 1}} \qquad |2 + \cos\theta + i\sin\theta|^2 = 5 + 4\cos\theta$

Watch out for (5)

Expanding the product to find its modulus
→ Modulus and Its Properties: |z₁z₂| = |z₁||z₂|
The argument from the ratio alone
→ The Argument: Reference Angle Plus the Quadrant
Applying De Moivre to sin θ + i cos θ
→ Polar Form and De Moivre: Powers, Rotations and sin θ + i cos θ
Answering the twin sitting's value
→ |z| + z = a + ib: Equate the Imaginary Part, Then Solve for |z|
z or z̄?
→ Find z From a Given Modulus: Simplify, Then Fix the Parameter

Test yourself on Complex Numbers

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