Question

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

Answer 1

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,

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