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4 years ago

Find the conjugate of the complex number, use division: 2+i —— 1+i

Mathematics

1 answer:

Fittoniya [83]4 years ago

4 0

I assume you mean divide using the conjugate to rationalize the denominator and express the result in standard rectangular form.

$\dfrac{2+i}{1+i} = \dfrac{2+i}{1+i} \cdot \dfrac{1-i}{1-i} = \dfrac{2(1)-i^2-2i+i}{1^2+1^2} = \frac 3 2 - \frac 1 2 i$

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Represent each of these relations on {1, 2, 3, 4} with a matrix (with the elements of this set listed in increasing order). a) {

Irina-Kira [14]

Answer:

for a a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

$\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]$

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

$\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]$

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

$\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]$

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

$\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]$

Step-by-step explanation:

in matrix, arrays are placed in rows , which represents the horizontal sides from left to right, while arrays in the column are placed vertically from top to bottom. Here, we placed the arrays in a 4x4 matrix

for a a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

$\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]$

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

$\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]$

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

$\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]$

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

$\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]$

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

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