2(z)^2 - 3(z')^2 = 10i: find all solutions

[courteous]

Find all solutions to equation \(\displaystyle 2(z)^2-3(\overline{z})^2=10i\)

I've put in for \(\displaystyle z=x+iy\) and got (after squaring): \(\displaystyle (2xy)i-(x^2-y^2)=10i \rightarrow x^2-y^2=0 \text{ and } xy=5 \rightarrow x^4=25\) Now what (I was given solution \(\displaystyle \{-1+i, -1+i\}\))?

Please, help me find a way to the solution.

Besides, I sense that my approach (inserting for \(\displaystyle z=x+iy\)) isn't the best one: which is (are) better?

 

[Deleted member 4993]

Re: 2(z)^2-3(z')^2=10i

Find all solutions to equation \(\displaystyle 2(z)^2-3(\overline{z})^2=10i\)

I've put in for \(\displaystyle z=x+iy\) and got (after squaring): \(\displaystyle (2xy)i-(x^2-y^2)=10i \rightarrow x^2-y^2=0 \text{ and } xy=5\)

\(\displaystyle \rightarrow x^4=25\)

Assuming you have done the arithmatic above correctly you have,

x[sup:48dpr1z8]4[/sup:48dpr1z8] - 5[sup:48dpr1z8]2[/sup:48dpr1z8] = 0

(x[sup:48dpr1z8]2[/sup:48dpr1z8] - 5)(x[sup:48dpr1z8]2[/sup:48dpr1z8] + 5) = 0

Now can you take these further.....
Now what (I was given solution \(\displaystyle \{-1+i, -1+i\}\))?

Please, help me find a way to the solution.

Besides, I sense that my approach (inserting for \(\displaystyle z=x+iy\)) isn't the best one: which is (are) better?

 

[DrMike]

I've put in for \(\displaystyle z=x+iy\) and got (after squaring): \(\displaystyle (2xy)i-(x^2-y^2)=10i \)

\(\displaystyle

This bit is incorrect. Check it again carefully.

Besides, I sense that my approach (inserting for ) isn't the best one

Well, you could let w = z[sup:3m7g8hir]2[/sup:3m7g8hir], and solve for w first (remembering that w' = z'[sup:3m7g8hir]2[/sup:3m7g8hir])\)