If (1 + i)(1 + 2i)(1 + 3i) dots (1 + ni) = a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given that $(1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni) = a + ib$ ..... $(i)$Taking the conjugate on both sides,we get $(1 - i)(1 - 2i)(1 - 3i) \dots (

Vedclass · Accepted Answer

$a^2 + b^2$

Vedclass · Answer

(B) Given that $(1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni) = a + ib$ ..... $(i)$Taking the conjugate on both sides,we get $(1 - i)(1 - 2i)(1 - 3i) \dots (1 - ni) = a - ib$ ..... $(ii)$Multiplying equation $(i)$ and equation $(ii)$,we get:$[(1 + i)(1 - i)] 	imes [(1 + 2i)(1 - 2i)] 	imes \dots 	imes [(1 + ni)(1 - ni)] = (a + ib)(a - ib)$Since $(1 + ki)(1 - ki) = 1^2 - (ki)^2 = 1 + k^2$,the expression becomes:$(1 + 1^2)(1 + 2^2)(1 + 3^2) \dots (1 + n^2) = a^2 + b^2$$2 	imes 5 	imes 10 	imes \dots 	imes (1 + n^2) = a^2 + b^2$

If $(1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni) = a + ib$,then $2 \times 5 \times 10 \times \dots \times (1 + n^2)$ is equal to

Similar Questions

If $\theta \in \mathbb{R}$ and $\frac{1-i \cos \theta}{1+2 i \cos \theta}$ is a real number,then $\theta$ will be (where $I$ is the set of integers):

If $x_n = \cos \left(\frac{\pi}{4^n}\right) + i \sin \left(\frac{\pi}{4^n}\right)$,then the product $x_1 x_2 x_3 \ldots \infty$ is equal to

The imaginary part of $\cosh(\alpha + i\beta)$ is

If $\cos (u + iv) = \alpha + i\beta ,$ then ${\alpha ^2} + {\beta ^2} + 1$ equals

If $i z^3+z^2-z+i=0$ (where $z$ is a complex number),then the value of $|z|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module