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

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The discussion focuses on evaluating complex powers of the imaginary unit $j$, specifically $j^{2312}$ and its reciprocal. It establishes that $j^2 = -1$, $j^3 = -j$, and $j^4 = 1$, which leads to the conclusion that $j^{2312} = 1$ since 2312 mod 4 equals 0. The reciprocal of $j^{2312}$ is also determined to be 1. The conversation emphasizes the cyclical nature of powers of $j$ and their reciprocals, highlighting a consistent pattern in their evaluations. Overall, the relationship between the initial evaluations and the final results is clarified through modular arithmetic.
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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!
 
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Drain Brain said:
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 $
 
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$.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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