4)For the following pairs (b, a) find y and x such that by + ax = d where d = GCD(a, b).

b)(b, a) = (18 - 16i, 7 - 2i) in the ring \(\displaystyle \mathbb{Z}\)

I started by dividing

\(\displaystyle \dfrac{18-16i}{7-2i}=\dfrac{158}{53}-\dfrac{76i}{53}\)

I then rounded each of these values to the nearest Gaussian integer to get 3 - i

\(\displaystyle (3-i)(7-2i)=19-13i \implies 18-16i=(3-i)(7-2i)-(1+3i)\)

That gives me the quotient and remainder to "feed" to the next round of division

\(\displaystyle \dfrac{7-2i}{-1-3i}=\dfrac{-1}{10}+\dfrac{23i}{10}\)

Rounding these values again gives me 0 + 2i

\(\displaystyle (0+2i)(-1-3i)=6-2i \implies 7-2i=(0+2i)(-1-3i)+1\)

Got another quotient and non-zero remainder, so I go again

\(\displaystyle \dfrac{-1-3i}{1}=1(-1-3i)+0\)

The remainder is zero, so I stop here. But now I have to build back up, to figure out what x and y are. I tried using the matrix method, but it fell apart:

\(\displaystyle \begin{pmatrix}1&1\\ 1&0\end{pmatrix} \cdot \begin{pmatrix}2i&1\\ 1&0\end{pmatrix} \cdot \begin{pmatrix}3-i&1\\ 1&0\end{pmatrix} \cdot \begin{pmatrix}0\\ 1\end{pmatrix} = \begin{pmatrix}1+2i\\ \:2i\end{pmatrix}\)

Ought to give me my values for x and y, but it's wrong for some reason.

\(\displaystyle (1+2i)(18-16i)+(2i)(7-2i)=54+34i\)

Instead of giving me the GCD(18-16i, 7-2i) = 1.

Every step I did matches up nicely with the in-class examples we worked... only the end result is wrong for this problem. The only thing I can maybe think of is that the examples we worked all involved only real numbers. Does this method just not work for complex numbers, or something? Any help or insight would be greatly appreciated.

I started by dividing

\(\displaystyle \dfrac{18-16i}{7-2i}=\dfrac{158}{53}-\dfrac{76i}{53}\)

I then rounded each of these values to the nearest Gaussian integer to get 3 - i

\(\displaystyle (3-i)(7-2i)=19-13i \implies 18-16i=(3-i)(7-2i)-(1+3i)\)

That gives me the quotient and remainder to "feed" to the next round of division

\(\displaystyle \dfrac{7-2i}{-1-3i}=\dfrac{-1}{10}+\dfrac{23i}{10}\)

Rounding these values again gives me 0 + 2i

\(\displaystyle (0+2i)(-1-3i)=6-2i \implies 7-2i=(0+2i)(-1-3i)+1\)

Got another quotient and non-zero remainder, so I go again

\(\displaystyle \dfrac{-1-3i}{1}=1(-1-3i)+0\)

The remainder is zero, so I stop here. But now I have to build back up, to figure out what x and y are. I tried using the matrix method, but it fell apart:

\(\displaystyle \begin{pmatrix}1&1\\ 1&0\end{pmatrix} \cdot \begin{pmatrix}2i&1\\ 1&0\end{pmatrix} \cdot \begin{pmatrix}3-i&1\\ 1&0\end{pmatrix} \cdot \begin{pmatrix}0\\ 1\end{pmatrix} = \begin{pmatrix}1+2i\\ \:2i\end{pmatrix}\)

Ought to give me my values for x and y, but it's wrong for some reason.

\(\displaystyle (1+2i)(18-16i)+(2i)(7-2i)=54+34i\)

Instead of giving me the GCD(18-16i, 7-2i) = 1.

Every step I did matches up nicely with the in-class examples we worked... only the end result is wrong for this problem. The only thing I can maybe think of is that the examples we worked all involved only real numbers. Does this method just not work for complex numbers, or something? Any help or insight would be greatly appreciated.