sketch

**scubasteve123** · January 11th 2010, 06:02 PM

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

**pickslides** · January 11th 2010, 06:08 PM

Originally Posted by scubasteve123

Re(z^4)=x^4 - 6x^2*y^2 + y^4

I agree with this.

**Prove It** · January 11th 2010, 06:09 PM

Originally Posted by scubasteve123

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

$z^4 = (x + iy)^4$

$= \sum_{k = 0}^4 {4\choose{k}}x^{4 - k}(iy)^k$

$= x^4 + 4ix^3y + 6i^2x^2y^2 + 4i^3xy^3 + i^4y^4$

$= x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4$

$= x^4 - 6x^2y^2 + y^4 + i(4x^3y - 4xy^3)$ .

So $\mathbf{Re}(z) = x^4 - 6x^2y^2 + y^4$

You are correct with your calculation.

Are you given any more information about $z$ ? Because you will need to evaluate $\mathbf{Re}(z) = \textrm{constant}$ in order to make the sketch...

**mr fantastic** · January 11th 2010, 06:10 PM

Originally Posted by scubasteve123

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

Your expression for Re(z^4) is correct. But there is no equation and therefore nothing to sketch unless Re(z^4) is equal to something ....

**scubasteve123** · January 11th 2010, 06:18 PM

Originally Posted by mr fantastic

Your expression for Re(z^4) is correct. But there is no equation and therefore nothing to sketch unless Re(z^4) is equal to something ....

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

**scubasteve123** · January 12th 2010, 06:33 PM

Im new to sketching in complex.
should i be trying to factor the expression I found for Re(z^4) ? or should i be able to sketch just using my expression? How do i know what this shape consists of? also since its only the real part is it just a line?

**Prove It** · January 12th 2010, 07:55 PM

Solve $x^4 - 6x^2y^2 + y^4 > 0$ for $y$ .

I would advise completing the square...

$y^4 - 6x^2y^2 + x^4 > 0$

$y^4 - 6x^2y^2 + (-3x^2)^2 - (-3x^2)^2 + x^4 > 0$

$(y^2 - 3x^2)^2 - 8x^4 > 0$

$(y^2 - 3x^2)^2 > 8x^4$

$|y^2 - 3x^2| > 2\sqrt{2}x^2$

$y^2 - 3x^2 < -2\sqrt{2}x^2$ or $y^2 - 3x^2 > 2\sqrt{2}x^2$

Can you go from here?

**scubasteve123** · January 13th 2010, 07:50 AM

would i now combine the x^2 terms then i get y^2>(2*2^1/2 +3)x^2
and y^2<(-2*2^1/2 +3)x^2 so now i see that the graph is greater than 2*2^1/2 +3 and less than -2*2^1/2 +3 ... im confused.

**Prove It** · January 13th 2010, 03:37 PM

Case 1:

$y^2 < (3 - 2\sqrt{2})x^2$

$|y| < x\sqrt{3 - 2\sqrt{2}}$

$x\sqrt{2\sqrt{2} - 3} < y < x\sqrt{3 - 2\sqrt{2}}$

Case 2:

$y^2 - 3x^2 > 2\sqrt{2}x^2$

$y^2 > (3 + 2\sqrt{2})x^2$

$|y| > \sqrt{(3 + 2\sqrt{2})x^2}$

$y <x\sqrt{- 3 - 2\sqrt{2}}$ or $y > x\sqrt{3 + 2\sqrt{2}}$ .

Now graph these inequalities.

**scubasteve123** · January 13th 2010, 04:12 PM

Originally Posted by Prove It

Case 1:

$y^2 < (3 - 2\sqrt{2})x^2$

$|y| < x\sqrt{3 - 2\sqrt{2}}$

$x\sqrt{2\sqrt{2} - 3} < y < x\sqrt{3 - 2\sqrt{2}}$

Case 2:

$y^2 - 3x^2 > 2\sqrt{2}x^2$

$y^2 > (3 + 2\sqrt{2})x^2$

$|y| > \sqrt{(3 + 2\sqrt{2})x^2}$

$y <x\sqrt{- 3 - 2\sqrt{2}}$ or $y > x\sqrt{3 + 2\sqrt{2}}$ .

Now graph these inequalities.

Im hoping that im finally having an "eureka" moment.. okay thank you for the help finding the inequalities... now to graph the inequalities do i just graph them like i would normally on the x,y plane since its only the Real part of the complex number??

**Prove It** · January 13th 2010, 04:51 PM

Originally Posted by scubasteve123

Im hoping that im finally having an "eureka" moment.. okay thank you for the help finding the inequalities... now to graph the inequalities do i just graph them like i would normally on the x,y plane since its only the Real part of the complex number??

Yes graph them as you normally would on the $x, y$ plane.

Just name your axes as $\mathbf{Re}(z)$ and $\mathbf{Im}(z)$ , since $x = \mathbf{Re}(z)$ and $y = \mathbf{Im}(z)$ .

**scubasteve123** · January 13th 2010, 05:03 PM

u've been so helpful. its really appreciated

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