Home » Ask & Discuss » Mathematics. » Analytical Geometry « Back to Discussion

Analytical Geometry

New kid on the Block

Joined:	15 Feb 2010
Post:	1

15 Feb 2010 04:21:12 IST

0 People liked this

430

who can solve the equation in the complex field of: (z^2)+2Re(z)=2

None

who can solve the equation in the complex field of:(z^2)+2Re(z)=2

Share this article on:

Comments (4)

SRIJAN MISHRA

Cool goIITian

Joined: 2 Feb 2010

Posts: 78

18 Feb 2010 21:54:11 IST

Like 1 people liked this

Its solution is -1+(3)^1/2 +0i

& -1-(3)^1/2 +0i

Yagyadutt Mishra

Forum Expert

Joined: 19 Feb 2009

Posts: 1984

18 Feb 2010 22:09:39 IST

Like 1 people liked this

let z be equal to x+iy

now z^2 = x^2 -y^2 +2.i.xy

x^2 -y^2 +2.x.y.i +2.x = 2

(x^2 -y^2 +2x -2) + 2xy.i = 0

so from here

xy = 0 ------(1)

and (x^2 -y^2 +2x -2) = 0 -----(2)

From (1) let x= 0 then from (2) y^2 = -2(which is not possible)

and again let y=0 then from (2) x^2 +2x -2 =0

(x+1)^2 -3 = 0

(x+1)^2 = 3

then x = root(3) -1 or -root(3) - 1

Hence z= x because we obtain x when y=0

So z = -1+root(3) or -1 -root(3)

Anant Kumar

Forum Expert

Joined: 10 Jul 2008

Posts: 598

18 Feb 2010 23:06:52 IST

Like 1 people liked this

Alternately,
The given equation is basically
$z^2+z+\bar{z}=2$ -----(1)
Taking conjugates
$\bar{z}^2+\bar{z}+z=2$ ---- (2)
From (1) and (2), we get $z^2=\bar{z}^2$ implying that either $z=\bar{z}$ or $z=-\bar{z}$
If $z=\bar{z}$ , then $z$ is purely real, so the given equation becomes
$z^2+2z-2=0\quad\Rightarrow \ z=-1\pm \sqrt{3}$
On the other hand, if $z=-\bar{z}$ , then $z$ is purely imaginary, i.e. $\mathrm{Re}(z)=0$ , so the original equation becomes
$z^2=2\quad\Rightarrow \ z=\pm \sqrt{2}$
But that's contradicting the fact that $z$ is purely imaginary. Hence, the only roots are $\boxed{z=-1\pm \sqrt{3}}$

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

19 Feb 2010 11:00:58 IST

Like 1 people liked this

z^2 = 2- 2Re(z) tells us that z^2 is real. That means that z is purely imaginary or z is real.

If z=ki, for some real number k, then the equation becomes k^2+2 = 0 which has no real roots.

So z is real and Re(z) = z.

and we have

$(z+1)^2= 3 \Rightarrow z+1 = \pm \sqrt 3 \Rightarrow z = -1 \pm \sqrt3$

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team