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.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 markMCQ
$\sqrt{5}+3i$
(a) $\dfrac{\sqrt{5}}{14}-\dfrac{3}{14}i$
(b) $\dfrac{\sqrt{5}}{14}+\dfrac{3}{14}i$
(c) $\dfrac{3}{14}-\dfrac{\sqrt{5}}{14}i$
(d) $\dfrac{3}{14}+\dfrac{\sqrt{5}}{14}i$
›Reveal solutionSolution

The multiplicative inverse of √5+3i is (√5−3i)/14.

For a complex number z = a+bi, its multiplicative inverse is $\dfrac{1}{z} = \dfrac{\bar z}{|z|^2} = \dfrac{a-bi}{a^2+b^2}$ .

Here a=√5, b=3, so $a^2+b^2 = 5+9=14$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 mark
$i^{-35}$
›Reveal solutionSolution

$i^{-35} = i$ , i.e., in the form a+ib this is 0+1i.

Powers of i cycle with period 4: $i^1=i, i^2=-1, i^3=-i, i^4=1$ , and this pattern repeats.

$i^{-35} = \dfrac{1}{i^{35}}$ . Since $35 = 4\times8+3$ , $i^{35}=i^{3}=-i$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 markMCQ
$z$
(a) $\dfrac{\bar{z}}{|z|^2}$
(b) $\dfrac{\bar{z}}{|z|}$
(c) $\dfrac{z}{|\bar{z}|}$
(d) $\dfrac{z}{|\bar{z}|^2}$
›Reveal solutionSolution

The multiplicative inverse of a nonzero complex number $z$ is $\dfrac{\bar z}{|z|^2}$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 markMCQ
$1 + i + i^2 + i^3 + i^4$
(a) $i$
(b) $0$
(c) $-i$
(d) $1$
›Reveal solutionSolution

$1+i+i^2+i^3+i^4 = 1$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 mark
$\left(\dfrac{1}{3} + 3i\right)^3$
›Reveal solutionSolution

$\left(\dfrac13+3i\right)^3 = -\dfrac{242}{27}-26i$ .

Use $(a+bi)^3 = (a^3-3ab^2) + i(3a^2b - b^3)$ with $a=\dfrac13$ , $b=3$ .

Real part: $a^3-3ab^2 = \left(\dfrac13\right)^3 - 3\cdot\dfrac13\cdot3^2 = \dfrac{1}{27} - 9 = \dfrac{1}{27}-\dfrac{243}{27} = -\dfrac{242}{27}$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL1 markMCQ
$\sqrt{i} + \sqrt{-i}$
(a) 0
(b) 1
(c) 2
(d) $\sqrt{2}$
›Reveal solutionSolution

Writing $i$ and $-i$ in polar form and taking their principal square roots, the imaginary parts cancel on addition, leaving $\sqrt{2}$ .

Write $i$ in polar (trigonometric) form: $i = \cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2}$ .

Step 1: Square root of $i$ .

By De Moivre's theorem, the principal square root is

$\sqrt{i} = \cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4} = \dfrac{\sqrt2}{2} + i\dfrac{\sqrt2}{2}$

Step 2: Square root of $-i$ .

Write $-i = \cos\left(-\dfrac{\pi}{2}\right) + i\sin\left(-\dfrac{\pi}{2}\right)$ , so its principal square root is …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Set ANNUAL1 mark
$(i^9 + i^{19})$
›Reveal solutionSolution

Reducing exponents mod $4$ gives $i^9=i$ and $i^{19}=-i$ , which cancel to give $0$ .

Since $i^4=1$ , powers of $i$ repeat every $4$ steps. Reduce each exponent modulo $4$ :

$i^9 = i^{4\times2+1} = (i^4)^2\cdot i = 1\cdot i = i$

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 markMCQ
$\sqrt{i} + \sqrt{-i}$
(a) 0
(b) 1
(c) 2
(d) $\sqrt{2}$
›Reveal solutionSolution

Using the principal square roots of $i$ and $-i$ in polar form, the sum simplifies to $\sqrt{2}$ .

$i$ has modulus $1$ and argument $\dfrac{\pi}{2}$ , so its principal square root is

$\sqrt{i}=\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}(1+i).$

$-i$ has modulus $1$ and argument $-\dfrac{\pi}{2}$ , so its principal square root is …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 mark
$3 - 4i$
›Reveal solutionSolution

Rationalising $\dfrac{1}{3-4i}$ using the conjugate $3+4i$ gives $\dfrac{3}{25}+\dfrac{4}{25}i$ .

The multiplicative inverse of $3-4i$ is $\dfrac{1}{3-4i}$ . Multiply numerator and denominator by the conjugate $3+4i$ :

$\dfrac{1}{3-4i}\times\dfrac{3+4i}{3+4i}=\dfrac{3+4i}{3^2+4^2}=\dfrac{3+4i}{9+16}=\dfrac{3+4i}{25}.$ …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 mark
$\dfrac{1}{1-i}$
›Reveal solutionSolution

Since $|1-i|=\sqrt2$ , the modulus of its reciprocal is $\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$ .

For any nonzero complex numbers, $\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}$ .

Here $z_1=1$ (modulus $1$ ) and $z_2=1-i$ , whose modulus is

$|1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}.$

So …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2020Set ANNUAL1 mark
$\sqrt{5} + 3i$
›Reveal solutionSolution

Multiply and divide by the conjugate to get the multiplicative inverse $\dfrac{\sqrt5-3i}{14}$ .

For a complex number $z = a+ib$ , its multiplicative inverse is

$z^{-1} = \dfrac{1}{z} = \dfrac{\bar z}{|z|^2} = \dfrac{a-ib}{a^2+b^2}$

Here $z = \sqrt5 + 3i$ , so $a=\sqrt5,\ b=3$ .

…

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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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