If (1 + i)(1 + 2i)(1 + 3i).....(1 + ni) = a + ib , then 2.5.10.... (1 + {n^2}) is equal to

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4-1.Complex numbers

medium

If $(1 + i)(1 + 2i)(1 + 3i).....(1 + ni) = a + ib$, then $2.5.10....$$(1 + {n^2})$ is equal to

${a^2} - {b^2}$

${a^2} + {b^2}$

$\sqrt {{a^2} + {b^2}} $

$\sqrt {{a^2} - {b^2}} $

Solution

(b)We have
$(1 + i)(1 + 2i)(1 + 3i)…..(1 + ni) = a + ib$ …..$(i)$
==> $(1 – i)(1 – 2i)(1 – 3i)…..(1 – ni) = a – ib$ …..$(ii)$
Multiplying $ (i)$ and $(ii)$, we get $2.5…..(1 + {n^2}) = {a^2} + {b^2}$

Standard 11

Mathematics

${\left\{ {\frac{{2i}}{{1 + i}}} \right\}^2} = $

easy

Let ${\left( { – \,2\, – \,\frac{1}{3}\,i} \right)^3} = \frac{{x \,+ \,iy}}{{27}}(i\, = \,\sqrt { – 1} ),$ where $x$ and $y$ are real numbers, then $y -x$ equals

hard

(JEE MAIN-2019)

Let ${z_1},{z_2}$ be two complex numbers such that ${z_1} + {z_2}$ and ${z_1}{z_2}$ both are real, then

medium

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normal

Additive inverse of $1 – i$is