Question

Solve :
$(i^{10}+1)(i^9+1)(i^8+1)........(i+1)$​

Answer 1

We have to use this concept.

$\iota^2=(\sqrt{-1})^2=-1\\ \\ \iota^6=\iota^{4+2}=\iota^4 \cdot \iota^2=\iota^2=-1\ \ \ \ \ \textsf{Since}\ \ \iota^{4n}=1\\ \\ \iota^{10}=\iota^{8+2}=\iota^8\cdot \iota^2=(\iota^4)^2\cdot \iota^2=\iota^2=-1\\ \\ \\ \large \therefore\ \iota^{4n+2}=-1\\ \\ \large \Longrightarrow\ \iota^{4n+2}+1=0$

So,

$\begin{aligned}&(\iota^{10}+1)(\iota^9+1)(\iota^8+1)\dots\dots(\iota+1)\\ \\ \Longrightarrow\ \ &(\iota^{4\cdot 2+2}+1)(\iota^9+1)(\iota^8+1)\dots\dots(\iota+1)\\ \\ \Longrightarrow\ \ &0(\iota^9+1)(\iota^8+1)\dots\dots(\iota+1)\\ \\ \Longrightarrow\ \ &\large \textbf{0}\end{aligned}$

Hence 0 is the answer.

$\begin{aligned}\Rightarrow\ \ &\textsf{In \ (\iota^{10}+1)(\iota^9+1)(\iota^8+1)\dots\dots(\iota+1), \ \ there are 3 factors}\\ &\textsf{ (\iota^{10}+1),\ (\iota^6+1),\ (\iota^2+1) \ \ which values 0 each.}\\ \\ \Rightarrow\ \ &(\iota+1)=(\iota^5+1)=(\iota^9+1)\\ \\ \Rightarrow\ \ &(\iota^3+1)=(\iota^7+1)=(-\iota+1)\ \ \ \ \ \textsf{since}\ \ \iota^{4n+3}=-\iota\\ \\ \Rightarrow\ \ &\iota^4+1=\iota^8+1=1+1=2\end{aligned}$

Answer 2

Answer:

Step-by-step explanation:

Complex numbers:-

Given a series.

We know that iota follows a cyclic series.

i^4n=1

i^(4n+1)=i

i^(4n+2)=-1

i^(4n+3)=-i

where n€W

So, putting n=2,we get i^10=-1

We find that the first term i^10+1 becomes 0.

Anything multiplied by 0 becomes 0

So,answer is 0