If $f(z)=\dfrac{7-z}{1-z^2}$, where $z=1+2i$, then $|f(z)|$ is: (A) $\dfrac{|z|}{2}$ (B) $|z|$ (C) $2|z|$ (D) none of these.

If f(z)=(7-z)/(1-z²), where z=1+2i, then |f(z)| is: (A) (|z|)/(2) (B) |z| (C) 2|z| (D) none of these.

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f(z)=\dfrac{7-z}{1-z^2}

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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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The problem asks for $|f(z)|$ when $z = 1+2i$ . The key is to substitute, simplify the complex fraction, find the modulus, and compare with $|z|$ . The final result is $|f(z)| = \frac{|z|}{2}$ , so option (A) is correct.

We start with the function $f(z) = \frac{7-z}{1-z^2}$ and $z = 1+2i$ . The goal is to compute $|f(z)|$ and match it with one of the given expressions in terms of $|z|$ .

First, recall that $|z|$ for $z = 1+2i$ is $\sqrt{1^2 + 2^2} = \sqrt{5}$ . But we won't need the numeric value directly — we'll work algebraically.

Why this approach works:

Instead of blindly plugging numbers, we simplify the expression step by step. Complex numbers often simplify nicely when you substitute and reduce. Here, $z^2$ will be a complex number, and the denominator $1 - z^2$ becomes something we can handle by rationalising or direct division.

Let’s go step by step.

Compute $z^2$

$z = 1 + 2i$

$z^2 = (1+2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i$
Find the denominator $1 - z^2$

$1 - z^2 = 1 - (-3 + 4i) = 1 + 3 - 4i = 4 - 4i$
Find the numerator $7 - z$

$7 - z = 7 - (1+2i) = 6 - 2i$

So now we have:

f(z) = \frac{6 - 2i}{4 - 4i}

Simplify the fraction Factor common terms: numerator = $2(3 - i)$ , denominator = $4(1 - i)$

f(z) = \frac{2(3 - i)}{4(1 - i)} = \frac{3 - i}{2(1 - i)}

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

f(z) = \frac{3 - i}{2(1 - i)} \cdot \frac{1 + i}{1 + i} = \frac{(3 - i)(1 + i)}{2(1 - i)(1 + i)}

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

Numerator: $(3 - i)(1 + i) = 3(1) + 3(i) - i(1) - i(i) = 3 + 3i - i - i^2 = 3 + 2i + 1 = 4 + 2i$

So:

f(z) = \frac{4 + 2i}{2 \cdot 2} = \frac{4 + 2i}{4} = 1 + \frac{i}{2}

Thus $f(z) = 1 + 0.5i$ .

Find $|f(z)|$ …

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Q.If f(z)=7−z1−z2f(z)=\dfrac{7-z}{1-z^2}f(z)=1−z27−z​, where z=1+2iz=1+2iz=1+2i, then ∣f(z)∣|f(z)|∣f(z)∣ is: (A) ∣z∣2\dfrac{|z|}{2}2∣z∣​ (B) ∣z∣|z|∣z∣ (C) 2∣z∣2|z|2∣z∣ (D) none of these.

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Q.If $f(z)=\dfrac{7-z}{1-z^2}$ , where $z=1+2i$ , then $|f(z)|$ is:
(A) $\dfrac{|z|}{2}$
(B) $|z|$
(C) $2|z|$
(D) none of these.

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