Order the following numbers from least to greatest

A complex number, represented by z = x + iy, may also be visualized as a 2 by 2 matrix

Marat540 [252]

Answer:

Step-by-step explanation:

A) Suppose that we have the complex numbers

$z= x + iy \quad \text{and} \quad \\\\ \tilde{z}=\tilde{x} + i \tilde{y}$

Remember that to sum complex numbers, we sum the real parts of the two numbers to get the real part and the imaginary parts of the two numbers to get the imaginary part. Hence,

$z+\tilde{z} = (x + i y) + (\tilde{x} + i \tilde{y}) = (x + \tilde{x})+i (y+\tilde{y})$

On the other hand, if we sum the matrix visualizations of $z \quad \text{and} \quad \tilde{z}$ we get

$\left[\begin{array}{cc}x &y\\-y&x\end{array}\right] + \left[\begin{array}{cc}\tilde{x}&\tilde{y}\\ -\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x + \tilde{x}& y + \tilde{y}\\-(y+\tilde{y})&x+\tilde{x}\end{array}\right]$

which is the matrix visualization of $z + \tilde{z}$ .

To multiply two complex numbers, we use the distributive law to multiplly and then separete the real part from the imaginary part

$z \cdot \tilde{z}= (x + iy) \cdot (\tilde{x} + i \tilde{y})=(x \tilde{x} + i x \tilde{y} + i \tilde{x} y - y\tilde{y} ) = (x\tilde{x}-y\yilde{y})+i(x\tilde{y}+\tilde{x}y)$

Again, if we multiply the matrix visualizations of $z \quad \text{and} \quad \tilde{z}$ we get

$\left[\begin{array}{cc}x&y\\-y&x\end{array}\right]\left[\begin{array}{cc}\tilde{x}&\tilde{y}\\-\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x\tilde{x}-y\tilde{y}&x\tilde{y}+y\tilde{x}\\-y\tilde{x}-x\tilde{y}&x\tilde{x}-y\tilde{y}\end{array}\right]$

which is the matrix viasualization of $z\cdot\tilde{z}.$

B) Since the usual matrix operations are consisten with the usual addition and multiplication rules in the complex numbers, we can use them to find the multiplicative inverses of a complex number $z=x+iy$ .

We are looking for the complex number $z^{-1}=(x+iy)^{-1}$ which in terms of matrices is equivalent to find the matrix

$\left[\begin{array}{cc}x&y\\-y & x\end{array}\right]^{-1}= \dfrac{1}{x^{2}+y^{2}} \left[\begin{array}{ccc}x&-y\\y&x\end{array}\right]$

Hence,

$z^{-1}=\dfrac{1}{x^2 +y^2} (x-iy)=\dfrac{1}{|z|^2}(x-iy)$

6 0

3 years ago

1) Lisa took a trip to Kuwait. upon leaving she decided to convert all of her dinars back into dollars . how many dollars did sh

Yuliya22 [10]

Lisa took a trip to Kuwait upon leaving she decided to convert all of her dinars back into dollars how many

3 0

3 years ago

Please important answer, as soon as possible I will thank you and mark as brainlist

denis23 [38]

10:12
15:18
i got this from doubling the first to make 5 = 10 and then that made 6 = 12. the second one was the same but tripled to make 6 = 18 and then that made 5 = 15

3 0

4 years ago

simplify -6/35 please

mojhsa [17]

Answer:

−0.171428571 or -6/35

Step-by-step explanation:

This is the Simplest form

6 0

2 years ago

The log of x cubed times y squared simplified

tresset_1 [31]