If $|z_1|=1$ ($z_1\neq-1$) and $z_2=\dfrac{z_1-1}{z_1+1}$, then show that the real part of $z_2$ is zero.

If |z₁|=1 (z₁≠-1) and z₂=(z₁-1)/(z₁+1), then show that the real part of z₂ is zero.

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|z_1|=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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To show the real part of $z_2$ is zero, we demonstrate that $z_2 = -\bar{z_2}$ . This is achieved by using the property that for $|z_1|=1$ , we have $\bar{z_1} = 1/z_1$ . The final result is that $\text{Re}(z_2) = 0$ .

When asked to show that the real part of a complex number is zero, we are essentially trying to prove that the number lies purely on the imaginary axis in the complex plane. For any complex number $z = x+iy$ , its real part is $x$ . We need to show $x=0$ .

A powerful way to do this is by using the relationship between a complex number and its conjugate.

z = x+iy

This approach often simplifies calculations significantly, especially when dealing with fractions of complex numbers.

The problem gives us a crucial piece of information: $|z_1|=1$ . This condition means $z_1$ is a complex number lying on the unit circle centered at the origin in the complex plane.

Important

For any complex number $z$ such that $|z|=1$ , a fundamental property is $z\bar{z} = |z|^2 = 1$ . This implies that $\bar{z} = \frac{1}{z}$ . We will use this property for $z_1$ .

The condition $z_1 \neq -1$ is important because it ensures that the denominator $z_1+1$ is not zero, so $z_2$ is well-defined.

Let's proceed with the steps:

Identify the expression for $z_2$ and its conjugate. We are given $z_2 = \frac{z_1-1}{z_1+1}$ . The conjugate of $z_2$ , denoted $\bar{z_2}$ , is found by taking the conjugate of each term in the expression:

\bar{z_2} = \overline{\left(\frac{z_1-1}{z_1+1}\right)} = \frac{\overline{z_1-1}}{\overline{z_1+1}} = \frac{\bar{z_1}-1}{\bar{z_1}+1}

Apply the property $\bar{z_1} = \frac{1}{z_1}$ . Since $|z_1|=1$ , we can substitute $\bar{z_1} = \frac{1}{z_1}$ into the expression for $\bar{z_2}$ :

\bar{z_2} = \frac{\frac{1}{z_1}-1}{\frac{1}{z_1}+1}

Simplify the expression for $\bar{z_2}$ . To simplify, we find a common denominator in the numerator and the denominator of the fraction:

\bar{z_2} = \frac{\frac{1-z_1}{z_1}}{\frac{1+z_1}{z_1}}

Now, we can cancel the $z_1$ terms from the numerator and denominator:

\bar{z_2} = \frac{1-z_1}{1+z_1}

Compare $z_2$ and $\bar{z_2}$ . We have $z_2 = \frac{z_1-1}{z_1+1}$ and $\bar{z_2} = \frac{1-z_1}{1+z_1}$ . Notice that the numerator of $\bar{z_2}$ is $-(z_1-1)$ . So, we can write $\bar{z_2}$ as: $\bar{z_2} = \frac{-(z_1-1)}{z_1+1} = -\left(\frac{z_1-1}{z_1+1}\right)$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQ
$(1 + i)(1 + i^2)(1 + i^3)(1 + i^4)$
›Reveal solutionSolution

$(1 + i^2) = 0$ , so the entire product equals $0$ .

Recall $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ .

So the second factor is $1 + i^2 = 1 - 1 = 0$ .

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2024Set ANNUAL1 mark
$z_1$
›Reveal solutionSolution

True — complex number multiplication is commutative, just like real number multiplication.

Let $z_1 = a+ib$ and $z_2 = c+id$ be any two complex numbers. Then

$z_1 z_2 = (a+ib)(c+id) = (ac-bd) + i(ad+bc)$

$z_2 z_1 = (c+id)(a+ib) = (ca-db) + i(cb+da) = (ac-bd)+i(ad+bc)$ …

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AHSEC Higher Secondary (HS) 1st Year Examination 2024Set ANNUAL1 mark
$k$
›Reveal solutionSolution

$i^{4k}=1$ for every integer $k$ .

Recall $i^1=i$ , $i^2=-1$ , $i^3=-i$ , $i^4=1$ , and then the pattern repeats every 4 powers since $i^4=1$ . So for any integer $k$ ,

$i^{4k} = (i^4)^k = 1^k = 1$ …

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AHSEC Higher Secondary (HS) 1st Year Examination 2023Set ANNUAL1 mark
$(i)^{2022}$
›Reveal solutionSolution

Powers of $i$ repeat every 4 steps, so reduce the exponent mod 4.

Recall $i^2 = -1$ , $i^3=-i$ , $i^4=1$ , and then the cycle repeats. So $i^n$ depends only on $n \bmod 4$ .

$2022 = 4 \times 505 + 2$ , so $2022 \equiv 2 \pmod 4$ .

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2022Set ANNUAL1 mark
$(i)^{-37}$
›Reveal solutionSolution

Powers of $i$ repeat with period $4$ : $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$ .

To evaluate $i^{-37}$ , reduce the exponent modulo $4$ . Since $-37 = 4(-10) + 3$ , we have $-37 \equiv 3 \pmod 4$ , so

$i^{-37} = i^{3} = -i$

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2022Set ANNUAL1 mark
$-i$
›Reveal solutionSolution

The multiplicative inverse (reciprocal) of a nonzero complex number $z$ is $1/z$ ; for a purely imaginary unit like $-i$ , rationalise by multiplying by its conjugate.

We need $(-i)^{-1} = \dfrac{1}{-i}$ .

Multiply numerator and denominator by $i$ (rationalising, since $i \cdot (-i)$ conveniently clears the imaginary denominator):

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2020Set ANNUAL1 mark
$(5-3i)^3$
›Reveal solutionSolution

Cube the binomial step by step; the result is $-10-198i$ .

We compute $(5-3i)^3$ in two stages.

Step 1 — square first:

$(5-3i)^2 = 25 - 2(5)(3i) + (3i)^2 = 25 - 30i + 9i^2 = 25 - 30i - 9 = 16 - 30i$

(using $i^2=-1$ ).

Step 2 — multiply by $(5-3i)$ once more:

$(5-3i)(16-30i) = 5(16) + 5(-30i) - 3i(16) - 3i(-30i)$ …

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AHSEC Higher Secondary (HS) 1st Year Examination 2018Set ANNUAL1 mark
$Q. Write the two complex cube roots of 1.$
›Reveal solutionSolution

The two complex (non-real) cube roots of $1$ are $\dfrac{-1+i\sqrt3}{2}$ and $\dfrac{-1-i\sqrt3}{2}$ .

We solve $x^3=1 \implies x^3-1=0 \implies (x-1)(x^2+x+1)=0$ .

One root is $x=1$ (real). The other two come from $x^2+x+1=0$ :

$x=\dfrac{-1\pm\sqrt{1-4}}{2}=\dfrac{-1\pm i\sqrt3}{2}.$ …

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Q.If $|z_1|=1$ ( $z_1\neq-1$ ) and $z_2=\dfrac{z_1-1}{z_1+1}$ , then show that the real part of $z_2$ is zero.

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