A number with the letter ” j ” in front of it identifies it as an imaginary number in the complex plane. By definition, the j-operator j ≡ √-1; Imaginary numbers can be added, subtracted, multiplied and divided the same as real numbers. The multiplication of ” j ” by ” j ” gives j 2 = -1

As an imaginary unit, use i or j (in electrical engineering), which satisfies the basic equation i 2 = −1 or j 2 = −1. The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle).

`(sqrta)^2 = a`, for any value of `a`. and `j = sqrt(-1)` Using these, we can derive the following: `j^2 = (sqrt-1)^2 = -1` Multiplying by `j` again gives us: `j^3 = j^2(j) = -j` Continuing the process gives us: `j^4 = j^3(j) = -j(j) = -(-1) = 1` `j^5 = j^4(j) = 1 × j = j` `j^6 = j^5(j) = j × j = …

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The J-2, commonly known as Rocketdyne J-2, was a liquid-fuel cryogenic rocket engine used on NASA's Saturn IB and Saturn V launch vehicles. Built in the United States by Rocketdyne, the J-2 burned cryogenic liquid hydrogen (LH 2) and liquid oxygen (LOX) propellants, with each engine producing 1,033.1 kN (232,250 lb f) of thrust in vacuum.

The Directorate for Intelligence, J-2, supports the Chairman of the Joint Chiefs of Staff, the Secretary of Defense, Joint Staff and Unified Commands. It is the national level focal point for crisis intelligence support to military operations, indications and warning intelligence in DoD, and Unified Command intelligence requirements.

Similar to the way that $i$ is $\sqrt{-1}$, but what that actually means is that $i^2$ is $-1$, $j^2$ is $+1$, but $j$ is not $1$. Here's the video with him saying all this - http://youtu.be/ygqIfLHGTu4?t=5m35s

Free math problem solver answers your algebra homework questions with step-by-step explanations.

Expand brackets as usual, but care with `j^2`! Example 2 - Multiplication . Multiply the following. a. `5(2 + 7j)` Answer

2018年4月19日 · $$J^{2}v_{4} = v_{2}; \quad J^{2}v_{2} = 0$$ One cycle has length 3 and the other has length 2, meaning the Jordan form has a block of size 3 and a block of size 2. Also, $\{v_1,v_3,v_5,v_2,v_4\}$ is a Jordan basis for $J^{2}$.