If $(1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy$, then $2\cdot5\cdot10\cdots(1+n^2)$ is (1) $1$\n(2) $i$\n(3) $x^2+y^2$\n(4) $1+n^2$

If (1+i)(1+2i)(1+3i)…(1+ni)=x+iy, then 2·5·10…(1+n²) is (1) 1n(2) in(3) x²+y²n(4) 1+n²

Exercise 2.9 · Q18

(1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy

(1)

1

\n(2)

i

\n(3)

x^2+y^2

\n(4)

1+n^2

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Concept understanding — Conjugate and Modulus

Conjugate. The conjugate of $z=x+iy$ is $\overline z=x-iy$ — obtained by flipping the sign of the imaginary part, equivalently by reflecting $z$ across the real axis in the Argand plane. A key fact: the product of a complex number with its own conjugate is always a non-negative real number, $z\overline z=(x+iy)(x-iy)=x^2+y^2$ .

Ten conjugate properties (each provable directly from the definition, several proved in the text):

$\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$
$\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}$
$\overline{z_1z_2}=\overline{z_1}\,\overline{z_2}$
$\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\overline{z_1}}{\overline{z_2}},\ z_2\ne0$
$\operatorname{Re}(z)=\dfrac{z+\overline z}{2}$
$\operatorname{Im}(z)=\dfrac{z-\overline z}{2i}$
$\overline{z^n}=(\overline z)^n$ , $n$ an integer
$z$ is real $\iff z=\overline z$
$z$ is purely imaginary $\iff z=-\overline z$
$\overline{\overline z}=z$

Proof idea (property 1): writing $z_1=x_1+iy_1,\ z_2=x_2+iy_2$ , $\overline{z_1+z_2}=(x_1+x_2)-i(y_1+y_2)=(x_1-iy_1)+(x_2-iy_2)=\overline{z_1}+\overline{z_2}$ . Proof idea (property 9): $z=-\overline z\iff x+iy=-(x-iy)=-x+iy\iff 2x=0\iff x=0$ , i.e. $z$ is purely imaginary.

The conjugate is the standard tool for dividing by a complex number: multiplying numerator and denominator by the conjugate of the denominator makes the denominator real (exactly like rationalising a surd).

Modulus. The modulus of $z=x+iy$ , written $|z|$ , is $|z|=\sqrt{x^2+y^2}$ — the distance from $z$ to the origin in the Argand plane, generalising the real-number absolute value. Note $z\overline z=|z|^2$ .

Eight modulus properties:

$|z|=|\overline z|$
$|z_1+z_2|\le|z_1|+|z_2|$ (Triangle Inequality)
$|z_1z_2|=|z_1||z_2|$
$\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}$
$|z_1-z_2|\ge\big||z_1|-|z_2|\big|,\ z_2\ne0$
$|z^n|=|z|^n$ , $n$ an integer
$\operatorname{Re}(z)\le|z|$
$\operatorname{Im}(z)\le|z|$ …

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Since $|z_1z_2\cdots z_n|=|z_1||z_2|\cdots|z_n|$ , taking moduli on both sides of the given product identity converts it into a statement about $2\cdot5\cdot10\cdots(1+n^2)$ versus $|x+iy|$ .

Step 1. Take the modulus of both sides of $(1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy$ .

|(1+i)(1+2i)\cdots(1+ni)|=|x+iy|.

Step 2. Use the modulus-of-a-product property.

|1+i|\cdot|1+2i|\cdot|1+3i|\cdots|1+ni|=|x+iy|.

Step 3. Compute each factor's modulus. For $1+ki$ , $|1+ki|=\sqrt{1+k^2}$ , so

\sqrt{1+1^2}\cdot\sqrt{1+2^2}\cdot\sqrt{1+3^2}\cdots\sqrt{1+n^2}=|x+iy|.

Step 4. Square both sides to clear the square roots. …

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

CBSE 2026Set ANNUAL1 markMCQ
$z$
(a) $2$
(b) $0$
(c) $3$
(d) $1$
›Reveal solutionSolution
Writing $z$ in polar form, the imaginary part of $z+1/z$ must vanish; since $z$ is non-real this forces $r=1/r$ , i.e. $|z|=1$ .

Let $z=r(\cos\theta+i\sin\theta)$ with $r=|z|>0$ . Since $z\in C\setminus R$ , $\sin\theta\ne0$ .

$\dfrac1z=\dfrac1r(\cos\theta-i\sin\theta)$ (since $\dfrac1{\cos\theta+i\sin\theta}=\cos\theta-i\sin\theta$ ).

$z+\dfrac1z=\left(r+\dfrac1r\right)\cos\theta+i\left(r-\dfrac1r\right)\sin\theta$ . …
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CBSE 2024Set ANNUAL1 markMCQ
$|z_1|=1$
(a) $3$
(b) $1$
(c) $4$
(d) $2$
›Reveal solutionSolution
Rewrites each conjugate as $\bar z_k=|z_k|^2/z_k$ to recognise the given expression as $z_1z_2z_3(\bar z_1+\bar z_2+\bar z_3)$ , then uses $|\bar w|=|w|$ .

Since $|z_1|=1$ : $\bar z_1=\dfrac{1}{z_1}$ . Since $|z_2|=2$ : $\bar z_2=\dfrac{4}{z_2}$ . Since $|z_3|=3$ : $\bar z_3=\dfrac{9}{z_3}$ .

Consider $z_1z_2z_3(\bar z_1+\bar z_2+\bar z_3)=z_1z_2z_3\left(\dfrac1{z_1}+\dfrac4{z_2}+\dfrac9{z_3}\right)=z_2z_3+4z_1z_3+9z_1z_2$ , which is exactly the given expression $9z_1z_2+4z_1z_3+z_2z_3$ .

So $9z_1z_2+4z_1z_3+z_2z_3=z_1z_2z_3(\bar z_1+\bar z_2+\bar z_3)$ , hence $|9z_1z_2+4z_1z_3+z_2z_3|=|z_1z_2z_3|\cdot|\bar z_1+\bar z_2+\bar z_3|$ . …
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CBSE 2024Set ANNUAL1 markMCQ
$(1+i)(1+2i)(1+3i)\ldots(1+ni)=x+iy$
(a) $x^2+y^2$
(b) $1$
(c) $1+n^2$
(d) $i$
›Reveal solutionSolution
Uses multiplicativity of the complex modulus on both sides of the given product, then squares to match $x^2+y^2$ .

Given $(1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy$ .

Take modulus of both sides: $|1+i|\,|1+2i|\cdots|1+ni|=|x+iy|$ .

Square both sides: $|1+i|^2|1+2i|^2\cdots|1+ni|^2=x^2+y^2$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$|z|=1$
(a) $\dfrac1z$
(b) $z$
(c) $1$
(d) $\bar{z}$
›Reveal solutionSolution
Using $|z|=1\Rightarrow\bar z=1/z$ turns the given ratio into a simple algebraic simplification.

Since $|z|=1$ , we have $z\bar z=|z|^2=1$ , so $\bar z=\dfrac1z$ . …
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CBSE 2022Set ANNUAL1 markMCQ
$(1+i)(1+2i)(1+3i)\ldots(1+ni)=x+iy$
(a) $x^2+y^2$
(b) $1$
(c) $1+n^2$
(d) $i$
›Reveal solutionSolution
Taking the modulus-squared of $(1+i)(1+2i)\cdots(1+ni)=x+iy$ turns the product of $(1+k^2)$ terms into $x^2+y^2$ .

We are given $(1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy$ .

Taking the modulus on both sides: $|1+i|\,|1+2i|\,|1+3i|\cdots|1+ni|=|x+iy|$ .

Since $|1+ki|=\sqrt{1+k^2}$ , the left side is $\sqrt{1+1^2}\sqrt{1+2^2}\sqrt{1+3^2}\cdots\sqrt{1+n^2}=\sqrt{2\cdot5\cdot10\cdots(1+n^2)}$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$(m-5)+i(n+4)$
(a) $\left(\dfrac{-1}{2}, -8\right)$
(b) $\left(\dfrac{-1}{2}, 8\right)$
(c) $\left(\dfrac{1}{2}, -8\right)$
(d) $\left(\dfrac{1}{2}, 8\right)$
›Reveal solutionSolution
Comparing real and imaginary parts after conjugating gives $m=-8$ and $n=-\dfrac12$ , i.e. $(n,m)=\left(-\dfrac12,-8\right)$ .

The complex conjugate of $(2m+3)+i(3n-2)$ is $(2m+3)-i(3n-2)$ .

Given: $(m-5)+i(n+4) = (2m+3)-i(3n-2)$ .

Equate real parts: $m-5 = 2m+3 \Rightarrow -m = 8 \Rightarrow m=-8$ . …
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Q.If (1+i)(1+2i)(1+3i)⋯(1+ni)=x+iy(1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy(1+i)(1+2i)(1+3i)⋯(1+ni)=x+iy, then 2⋅5⋅10⋯(1+n2)2\cdot5\cdot10\cdots(1+n^2)2⋅5⋅10⋯(1+n2) is

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Q.If $(1+i)(1+2i)(1+3i)\cdots(1+ni)=x+iy$ , then $2\cdot5\cdot10\cdots(1+n^2)$ is

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