Sketch Re(z^4). ie:sketch real part of z^4
I tried to do the arithmetic and I think Re(z^4)=x^4 - 6x^2*y^2 + y^4
$\displaystyle z^4 = (x + iy)^4$
$\displaystyle = \sum_{k = 0}^4 {4\choose{k}}x^{4 - k}(iy)^k$
$\displaystyle = x^4 + 4ix^3y + 6i^2x^2y^2 + 4i^3xy^3 + i^4y^4$
$\displaystyle = x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4$
$\displaystyle = x^4 - 6x^2y^2 + y^4 + i(4x^3y - 4xy^3)$.
So $\displaystyle \mathbf{Re}(z) = x^4 - 6x^2y^2 + y^4$
You are correct with your calculation.
Are you given any more information about $\displaystyle z$? Because you will need to evaluate $\displaystyle \mathbf{Re}(z) = \textrm{constant}$ in order to make the sketch...
Hi everyone thanks for checking I was so conserned with my arithmetic i forgot the rest my apologies. I need to sketch {z element complex numbers such that Re(z^4)>0 and |z-1|<2}
I know how to sketch the |z-1|<2 part and i assume the part I want is where the 2 overlap i just not sure how to sketch x^4-6x^2*y^2 +y^4
Solve $\displaystyle x^4 - 6x^2y^2 + y^4 > 0$ for $\displaystyle y$.
I would advise completing the square...
$\displaystyle y^4 - 6x^2y^2 + x^4 > 0$
$\displaystyle y^4 - 6x^2y^2 + (-3x^2)^2 - (-3x^2)^2 + x^4 > 0$
$\displaystyle (y^2 - 3x^2)^2 - 8x^4 > 0$
$\displaystyle (y^2 - 3x^2)^2 > 8x^4$
$\displaystyle |y^2 - 3x^2| > 2\sqrt{2}x^2$
$\displaystyle y^2 - 3x^2 < -2\sqrt{2}x^2$ or $\displaystyle y^2 - 3x^2 > 2\sqrt{2}x^2$
Can you go from here?
Case 1:
$\displaystyle y^2 < (3 - 2\sqrt{2})x^2$
$\displaystyle |y| < x\sqrt{3 - 2\sqrt{2}}$
$\displaystyle x\sqrt{2\sqrt{2} - 3} < y < x\sqrt{3 - 2\sqrt{2}}$
Case 2:
$\displaystyle y^2 - 3x^2 > 2\sqrt{2}x^2$
$\displaystyle y^2 > (3 + 2\sqrt{2})x^2$
$\displaystyle |y| > \sqrt{(3 + 2\sqrt{2})x^2}$
$\displaystyle y <x\sqrt{- 3 - 2\sqrt{2}}$ or $\displaystyle y > x\sqrt{3 + 2\sqrt{2}}$.
Now graph these inequalities.