Complex number 4

Let's solve that system only for $x$ by keeping fixed $-\dfrac{y}{x^2+y^2}$ and $\dfrac{x}{x^2+y^2}$ :

$x=\dfrac{33x-56y}{4225(x^2+y^2)}$

Now, take the second equation and substitute it:

$x=\dfrac{\dfrac{x}{x^2+y^2}}{4225(x^2+y^2)}$

Since $x \neq 0$ :

$4225(x^2+y^2)^2=1 \\ x^2+y^2= \pm \dfrac{1}{65}$

Now, substitute in the first equation we obtained:

$x=\dfrac{33x-56y}{\pm 65} \\ \pm 65x=33x-56y \\ x(33\mp 65)=56y \\ x=\dfrac{56}{33\mp 65}y$

First consider the first choice of sign, so we obtain:

$x=-\dfrac{7}{4}y$

Substitute in $x^2+y^2=\dfrac{1}{65}$ :

$\dfrac{49}{16}y^2+y^2=\dfrac{1}{65} \\ y=\pm\dfrac{4}{65} \\ x=\mp \dfrac{7}{65}$

In this case we obtain the solutions: $(x,y)=\left(\mp \dfrac{7}{65},\pm\dfrac{4}{65}\right)$

Finally, consider the second choice of sign:

$x=\dfrac{4}{7}y$

Substitute in $x^2+y^2=-\dfrac{1}{65}$ :

$\dfrac{16}{49}y^2+y^2=-\dfrac{1}{65} \\ y=\pm \dfrac{7}{65}i \\ x=\pm \dfrac{4}{65}i$

In this case we obtain the solutions: $(x,y)=\left(\pm \dfrac{4}{65}i,\pm\dfrac{7}{65}i\right)$

In both cases, we have that $|x|+|y|=\dfrac{4}{65}+\dfrac{7}{65}=\dfrac{11}{65}$ , so $p=11$ , $q=65$ and $6p-q=\boxed{1}$ .

Let $z=x+iy,where \quad i=\sqrt{-1}$

We can write the given system as $z(33+56i)=\frac {\bar{z}}{|z|^2} \implies z(33+56i)=\frac{1}{z}$ (Using $z\bar{z}=|z|^2$ )

$\implies z^2(33+56i)=1$

$\implies z^2=\frac{1}{33+56i}$

$\implies z^2=\frac{33-56i}{33^2+56^2}$

$\implies z^2=\frac{33-56i}{65^2}$

Comparing imaginary and real parts of $z^2$

$\implies \begin{cases} x^2-y^2=\frac{33}{65^2} \\ 2xy=\frac{-56}{65^2} \end{cases}$

Eliminating $x$ we get

$\quad 4y^2(\frac{33}{65^2}+y^2)=\frac{56^2}{65^4}\\ \implies 4y^4+4y^2 \frac{33}{65^2}-\frac{56^2}{65^4}=0\\ \implies y^4+y^2\frac{33}{65^2}-\frac{56^2}{4 \cdot 65^4}=0\\ \implies(y^2+\frac{33}{2 \cdot 65^2})^2=\frac{1}{4 \cdot 65^2}\\ \implies (y^2+\frac{33}{2 \cdot 65^2})=\frac{1}{2 \times 65}\\ \implies y^2=\frac{32}{2 \cdot 65^2}\\ \implies |y|=\frac{4}{65} \quad |x|=\frac{7}{65}$

$|x|+|y|=\frac{11}{65} \\ \boxed{p=11} \quad \boxed{q=65}\\ \boxed{6p-q=1}$

We square both equations in the given system to obtain

$\begin{aligned} 56^2x^2 + (2\times56\times33)\,xy + 33^2y^2 &= \dfrac{y^2}{(x^2+y^2)^2}\\ 33^2x^2 - (2\times33\times56)\,xy + 56^2y^2 &= \dfrac{x^2}{(x^2+y^2)^2} \end{aligned}$

Adding these two resulting equations, we get

$\begin{aligned} (56^2+33^2) (x^2+y^2) &= \dfrac{x^2+y^2}{(x^2+y^2)^2}\\ 65^2(x^2+y^2) &= \dfrac{1}{x^2+y^2}\\ x^2+y^2 &= \pm \dfrac{1}{65} \end{aligned}$

Combining the last line with the original system leads to the pair of systems

$\begin{aligned} 56x + 33y &= \mp 65y\\ 33x - 56y &= \pm 65x \end{aligned}$

Unfortunately neither of these two systems is independent; solving them leads to $4x = -7y$ and $7x = 4y$ respectively. However, squaring these equations and then combining them with the earlier obtained $x^2+y^2 = \dfrac{1}{65}$ and $x^2+y^2 = -\dfrac{1}{65}$ respectively yields results.

First,

$\begin{aligned} 16x^2 &= 49 y^2\\ x^2 + y^2 &= \dfrac{1}{65} \end{aligned}$

leads to $x^2 = \dfrac{49}{65^2} \text{ and } y^2 = \dfrac{16}{65^2} \text{, so } |x| = \dfrac{7}{65} \text{ and } |y| = \dfrac{4}{65}.$

Second,

$\begin{aligned} 49x^2 &= 16 y^2\\ x^2 + y^2 &= -\dfrac{1}{65} \end{aligned}$

leads to $x^2 = -\dfrac{16}{65^2} \text{ and } y^2 = -\dfrac{49}{65^2} \text{, so } |x| = \dfrac{4}{65} \text{ and } |y| = \dfrac{7}{65}.$

In both cases, we get $|x| + |y| = \dfrac{11}{65}\text{, so } p=11 \text{ and } q=65 \text{, thus } 6p-q= \boxed{1}$ .