If $|z_1|=|z_2|=\ldots=|z_n|=1$, then show that $|z_1+z_2+z_3+\ldots+z_n|=\left|\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}+\ldots+\dfrac{1}{z_n}\right|$.

If |z₁|=|z₂|=…=|zₙ|=1, then show that |z₁+z₂+z₃+…+zₙ|=|(1)/(z₁)+(1)/(z₂)+(1)/(z₃)+…+(1)/(zₙ)|.

NCERT Exemplar · Q20

|z_1|=|z_2|=\ldots=|z_n|=1

Arunachal CbseLong· 3mImportance★★★★★est

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Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$ . You know that no real number squared gives $-1$ . The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$ ? That's exactly what mathematicians did — and that invention is the imaginary unit $i$ , defined by:

i^2 = -1

A complex number is any number of the form $a + bi$ , where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$ .

Note

The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$ .

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$ .

(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Watch out

The most common mistake: forgetting that $i^2 = -1$ , not $1$ . Always check your final step.

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$ . When you multiply a complex number by its conjugate, you get a real number:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

So to divide:

\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

Tip

To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes $c^2 + d^2$ , a positive real number.

The Big Picture …

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For complex numbers on the unit circle, each number equals the conjugate of its reciprocal. This identity lets us rewrite the sum of reciprocals as the conjugate of the original sum, and since magnitude is unchanged by conjugation, the two sums have equal absolute value.

The core idea here is beautifully simple once you see it. When a complex number lies on the unit circle — that is, $|z| = 1$ — it has a special property: its reciprocal is exactly its complex conjugate. Let's see why.

If $|z| = 1$ , then $z \overline{z} = |z|^2 = 1$ , so $\overline{z} = \frac{1}{z}$ . This is the key that unlocks the entire problem.

Now, we are given $n$ complex numbers $z_1, z_2, \ldots, z_n$ , each with $|z_k| = 1$ . We want to compare the magnitude of their sum with the magnitude of the sum of their reciprocals.

Apply the unit-circle property to each term.

Since $|z_k| = 1$ for every $k$ , we have $\frac{1}{z_k} = \overline{z_k}$ for each $k = 1, 2, \ldots, n$ .
Rewrite the sum of reciprocals.

\frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_n} = \overline{z_1} + \overline{z_2} + \cdots + \overline{z_n}

Use the fact that the conjugate of a sum is the sum of the conjugates.

\overline{z_1} + \overline{z_2} + \cdots + \overline{z_n} = \overline{z_1 + z_2 + \cdots + z_n}

Now compare magnitudes. The magnitude of a complex number and the magnitude of its conjugate are always equal: $|\overline{w}| = |w|$ for any complex number $w$ . Therefore, $\left| \frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_n} \right| = \left| \overline{z_1 + z_2 + \cdots + z_n} \right| …$

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