Powers of Imaginary Unit i and Complex Numbers - Class XII

$\left| \begin{matrix} 2+i & 2-i \ 1+i & 1-i \end{matrix} \right| \ =(2+i)(1-i)-(2+i)(1+i)\ =2-2i+i-{ i }^{ 2 }-2-2i-i-{ i }^{ 2 }\ =-4i+2$

$(i)^{242}$

$=(i^{2})^{121}$

$=(-1)^{121}$                              since $i=\sqrt{-1}$.
Hence
$(-1)^{121}=-1$
Thus
$i^{242}=-1$

Let $Z= i+\dfrac{1}{i}$

$(-\sqrt{-1})^{4n+3} = (-i)^{4n+3}
= (-i)^{4n}(-i)^3
= {(-i)^4}^n(-i)^3
= 1 \times (-i)^3 = i$

The powers of i repeat every 4: i^1=i, i^2=-1, i^3=-i, i^4=1. Divide 135 by 4: 135 = 4 * 33 + 3. Thus, i^135 = i^3 = -i.

The value of $(i)^{457} $

$\displaystyle \left ( \frac{1 + i}{1 - i} \right )^n = 1$         ${ \because -i = \displaystyle \Rightarrow \frac{1}{i}}$
$\displaystyle \Rightarrow\left ( \frac{1 + i}{\displaystyle 1 + \frac{1}{i}} \right )^n = 1$
$i^n = 1$
so min value of $n =4$

$\left(\dfrac{1+i}{1-i}\right)^2+\left(\dfrac{1-i}{1+i}\right)^2=\left[\dfrac{(1+i)(1+i)}{(1-i)(1+i)}\right]^2+\left[\dfrac{(1-i)(1-i)}{(1+i)(1-i)}\right]^2$

$\sqrt {-1} = i$ $(iota)$

$-3\sqrt {-10}$ $= -3\sqrt {-1} \times \sqrt {10}$

After expanding, we get $(4+2i)(4-2i)=16+8i-8i-4i^4$

$i$ is an imaginary number whose value is $\sqrt { -1 } $

Given:

$ i^{24} + (\cfrac{1}{i})^{26}$

${i}^{57}-(\cfrac{1}{{i}^{25}})$

The cube roots of i can be found using De Moivre's Theorem. i = cos(pi/2) + i sin(pi/2). The roots are cos((pi/2 + 2kpi)/3) + i sin((pi/2 + 2kpi)/3) for k=0, 1, 2. For k=0, we get cos(pi/6) + i sin(pi/6) = sqrt(3)/2 + i/2.

$z+\cfrac{1}{z}=2\cos {6}^{o}$

$(1+i)^4+(1-i)^4$

Simplify the terms: (1+i)^n1 + (1-i)^n1 + (1+i)^n2 + (1-i)^n2. Since (1+i) and (1-i) are conjugates, their sum to any power n is real. Thus, the entire expression is real.

Given:-

$\triangle =\left| \begin{matrix} { a } _{ 11 }\quad \quad  & { a } _{ 12 }\quad \quad  & { a } _{ 13 } \ { a } _{ 21 }\quad \quad  & { a } _{ 22 }\quad \quad  & { a } _{ 23 } \ { a } _{ 31 }\quad \quad  & { a } _{ 32 }\quad \quad  & { a } _{ 33 } \end{matrix} \right| \quad & \quad { a } _{ pq }={ i }^{ p+q }$

$\Rightarrow \quad \triangle =\left| \begin{matrix} { i }^{ 2 }\quad \quad  & { i }^{ 3 }\quad \quad  & { i }^{ 4 } \ { i }^{ 3 }\quad \quad  & { i }^{ 4 }\quad \quad  & { i }^{ 5 } \ { i }^{ 4 }\quad \quad  & { i }^{ 5 }\quad \quad  & { i }^{ 6 } \end{matrix} \right| ={ i }^{ 2+3+4 }\left| \begin{matrix} { 1 }\quad \quad  & 1\quad \quad  & 1 \ { i }\quad \quad  & { i }\quad \quad  & { i } \ { i }^{ 2 }\quad \quad  & { i }^{ 2 }\quad \quad  & { i }^{ 2 } \end{matrix} \right| $


$=i\left| \begin{matrix} 1 & 1 & 1 \ i & i & i \ -1 & -1 & -1 \end{matrix} \right| =-i\left| \begin{matrix} 1 & 1 & 1 \ i & i & i \ 1 & 1 & 1 \end{matrix} \right| $

$\therefore \quad \triangle =0$
Hence, option 'C' is correct.

$S=i+2i^2+3i^3\ldots+100i^{100}\S=i-2-3i+4+\ldots +100\S=i(1-3+5\ldots)+(-2+4\ldots)\S=50-50i=50(1-i) $

Let $Z=\dfrac{i^6+i^7+i^8+i^9}{i^2+i^3}$

We have $i^2 = -1$
Thus, $\displaystyle \sum _{n=1}^{4}(i^n+i^{n+1}) = (i^1 + i^2) + (i^2 + i^3) + (i^3 + i^4) + (i^4 + i^5) = (i - 1) + (-1 - i) + (-i + 1) + (1 + i) = 0$
\Rightarrow $\displaystyle \sum _{n=1}^{12}(i^n+i^{n+1}) = 0$
Now, only remains is $i^{13} + i^{14} = i - 1$

$5\sqrt {-8}$ $= 5\sqrt {8}\times \sqrt {-1}$

$2\sqrt {-49}$ $= 2\sqrt {-1 \times 7\times 7}$

$\sqrt {-36}$ $=\sqrt {-1 \times 6 \times 6}$

${ i }^{ 413 }{ i }^{ x }=1$

Consider $(4-2i)(-3+3i)$

Given, $i=\sqrt {-1}$

We have,

$ {{\left( \dfrac{2i}{1+i} \right)}^{2}} $

$ \Rightarrow \dfrac{4{{i}^{2}}}{1+{{i}^{2}}+2i} $

$ \Rightarrow \dfrac{-4}{1-1+2i}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because {{i}^{2}}=-1 \right) $

$ \Rightarrow \dfrac{-4}{2i} $

$ \Rightarrow \dfrac{-2}{i} $

$ \Rightarrow \dfrac{-2i}{{{i}^{2}}} $

$ \Rightarrow \dfrac{-2i}{-1} $

$ \Rightarrow 2i $

Hence, this is the answer.

$\because \dfrac {1 + i}{1 - i} = \dfrac {1 + i}{1 - i}\times \dfrac {1 + i}{1 + i}$
$= \dfrac {(1 + i)^{2}}{1 - i^{2}} = \dfrac {1 + 2i + i^{2}}{1 - i^{2}}$
$= \dfrac {1 + 2i - 1}{1 + 1} = i$
$\therefore \left (\dfrac {1 + i}{1 - i}\right )^{n} = 1$
$\Rightarrow i^{n} = 1$
Thus, $i^{n}$ will be positive integer, if $n = 4$.

For $n$ is positive integer, $4n+1$ is an odd integer$(5,9,13,....)$

$(1+i)^{n _{1}}+(1+i^{2})^{n _{2}}+(1+i^{5})^{n _{2}}+(1+i^{7})^{n _{2}}$
$=(1+i)^{n _{1}}+(1-i)^{n _{2}}+(1+i)^{n _{2}}+(1-i)^{n _{2}}$
$=2[1+:^{n _{1}}C _{2}i^{2}+:^{n _{1}}C _{4}i^{4}...]+2[1+:^{n _{2}}C _{2}i^{2}+:^{n _{2}}C _{4}i^{4}...]$
$=2[1+-:^{n _{1}}C _{2}+:^{n _{1}}C _{4}-...]+2[1-:^{n _{2}}C _{2}+:^{n _{2}}C _{4}-...]$
Hence
For all $n _{1}>0$ and $n _{2}>0$ the above expression yields real integral number.
Where $n _{1},n _{2}\epsilon N$.
Hence, option 'D' is correct.

$\displaystyle \sum _{n = 1}^{11} (i^{n} + i^{n + 1}) i = \sqrt {-1}$