MATH!1!1!!

x+(y*y)=5

(z^y)-z=3

What is z

3x = 18 x = 6

6 + y^2 = 5 y^2 = -1 y = i

can be written as

e^(i*ln(z))

and that in turn as

e^(i*ln(|z|)- arg(z))

and then as

e^(-arg(z)) (cos(ln|z|) + i*sin(ln|z|))

so the last equation can be rewritten as:

e^(-arg(z))*cos(ln|z|)+i* e^(-arg(z))*sin(ln|z|) - z = 3

From here we can seperate imaginary and real parts:

e^(-arg(z)) *sin(ln|z|) = Im(z)

e^(-arg(z))*cos(ln|z|) = 3 + Re(z)

then:

-arg(z) = -arctan(Im(z)/Re(z))

and

z = a + i*b

which means

Re(z) = a

Im(z) = b

by definition.

so we got:

sin(ln(a^2+b^2)/2)/(e^(arctan(b/a)) = b

cos(ln(a^2+b^2)/2)/(e^(arctan(b/a)) = 3 + a

This is a system of equations with 2 variables and 2 equations so it should have a solution, but I belive this to be either beyond my ability or analitically unsolvable. It is ofc possible the solution does not exists or there is infinitely many (especially since trig is involved).

If someone else wants to take a swing at numerically finding the solution go ahead. I wont even ask WolframAlpha for anything for I am too lazy for that.

So, cos(x+pi/2)=sin(x), so we'll convert the sin from the first equation into cos using that

Move everything to the otherside

And then substitute one into the other

Now, arctan can be rewritten in it's logarithmic form to help remove e

Doing the same with arccos gives us

Now, we can divide all of the elements by -i

Then turn -pi/2i into a logarithm

And now we can finally put the two logarithms together using the log(a)+log(b)=log(ab) rule

And since e^pi/2i=i

We can further simplify the equation

Chaotic Good: Two points (ordered) define a plane by describing its normal vector