How can $j^{2}$ and $j^{2312}$ be related in complex number evaluations?

[Drain Brain]

can you help me how to formally present a solution to this problem

show that $j^2=-1$, $j^3=-j$, $j^4=1$ and $\frac{1}{j}=-j$, $\frac{1}{j^2}=-1$, $\frac{1}{j^3}=-j$, $\frac{1}{j^4}=1$. From the result of these, evaluate $j^{2312}$ and its reciprocal.

sure I can solve for $j^{2312}$ and its reciprocal, but the thing that I need help with is how can I relate the first part of the task to the last part of the problem given.

thanks!

 

[kaliprasad]

can you help me how to formally present a solution to this problem

show that $j^2=-1$, $j^3=-j$, $j^4=1$ and $\frac{1}{j}=-j$, $\frac{1}{j^2}=-1$, $\frac{1}{j^3}=-j$, $\frac{1}{j^4}=1$. From the result of these, evaluate $j^{2312}$ and its reciprocal.

sure I can solve for $j^{2312}$ and its reciprocal, but the thing that I need help with is how can I relate the first part of the task to the last part of the problem given.

thanks!

you know $j^4=1$
2312 mod 4 = 0

hence $j^{2312} =1$

its reciprocal = $\frac{1}{1} = 1 $

 

[Deveno]

The powers of $i$ form a 4 cycle:$1 \to i \to -1 \to -i \to 1 \to \cdots$

The reciprocals (multiplicative inverses) of powers of $i$ form the same cycle, IN REVERSE ORDER:

$1 =\dfrac{1}{1} \to -i = \dfrac{1}{i} \to -1 = \dfrac{1}{i^2} \to i = \dfrac{1}{i^3} \to 1 \cdots$

Suppose we want to find: $\dfrac{1}{i^n}$

and we know that $n = 4k + 3$.

then: $\dfrac{1}{i^n} = \dfrac{1}{i^{4k+3}} = \dfrac{1}{i^{4k}}\cdot\dfrac{1}{i^3}$$= \dfrac{1}{(i^4)^k}\cdot\dfrac{1}{i^3} = \dfrac{1}{1^k}\cdot\dfrac{1}{i^3}$$= 1\cdot\dfrac{1}{i^3} = \dfrac{1}{i^3} = \dfrac{i^4}{i^3} = i$.