If $\dfrac{z-1}{z+1}$ is a purely imaginary number ($z\neq-1$), then find the value of $|z|$.

If (z-1)/(z+1) is a purely imaginary number (z≠-1), then find the value of |z|.

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\dfrac{z-1}{z+1}

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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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A complex number is purely imaginary when its real part is zero. Using this condition on $\frac{z-1}{z+1}$ forces $z$ to lie on the perpendicular bisector of the segment joining $1$ and $-1$ in the complex plane, which is the imaginary axis. The result: $|z| = 1$ .

When a fraction of complex numbers is purely imaginary, we're saying it has no real component—it sits entirely on the imaginary axis. This geometric constraint on $\frac{z-1}{z+1}$ will translate into a powerful algebraic condition on $z$ itself.

The standard approach is to write $w = \frac{z-1}{z+1}$ and demand that $w$ is purely imaginary, meaning $w = -\overline{w}$ . (A number equals the negative of its conjugate precisely when its real part vanishes.) Let's apply this systematically.

Solution

Set up the purely imaginary condition

Let $w = \frac{z-1}{z+1}$ . For $w$ to be purely imaginary, we need $w + \overline{w} = 0$ , which is equivalent to $w = -\overline{w}$ .

Taking the conjugate of both sides:

\overline{w} = \overline{\left(\frac{z-1}{z+1}\right)} = \frac{\overline{z}-1}{\overline{z}+1}

Apply the condition $w = -\overline{w}$

Substituting:

\frac{z-1}{z+1} = -\frac{\overline{z}-1}{\overline{z}+1}

Cross-multiplying:

(z-1)(\overline{z}+1) = -(z+1)(\overline{z}-1)

Expand both sides

Left side:

z\overline{z} + z - \overline{z} - 1

Right side:

-(z\overline{z} - z + \overline{z} - 1) = -z\overline{z} + z - \overline{z} + 1

Equate and simplify …

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Q.If z−1z+1\dfrac{z-1}{z+1}z+1z−1​ is a purely imaginary number (z≠−1z\neq-1z=−1), then find the value of ∣z∣|z|∣z∣.

Solution

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Q.If $\dfrac{z-1}{z+1}$ is a purely imaginary number ( $z\neq-1$ ), then find the value of $|z|$ .

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