Is the Complex Equation z^2 - z |z| = 1 Solvable?

[Quinzio]

Complex equation with "z"

Homework Statement

$$z^2-\bar z |z|= 1$$

Homework Equations

Complex numbers

The Attempt at a Solution

$$(a+ib)^2-(a-ib)\sqrt{a^2+b^2} = 1$$

$$a^2-b^2+2iab-a\sqrt{a^2+b^2}+ib\sqrt{a^2+b^2} = 1$$

I separate the real part from the imaginary part.

$$\left\{\begin{matrix} 2iab+ib\sqrt{a^2+b^2}=0 \\ a^2-b^2-a\sqrt{a^2+b^2}=1 \end{matrix}\right.$$

One solution of the first is $b=0$

$$\left\{\begin{matrix} 2a+\sqrt{a^2+b^2}=0, \ 3a^2=b^2 \\ a^2-b^2-a\sqrt{a^2+b^2}=1 \end{matrix}\right.$$

So I basicalliy have two solution from the first eq.
1) $b=0$
2)$3a^2=b^2$

If you plug these results in the seconds I don't get anything...
Either I get $a^2-a^2=1$ or $-a^2=1$ which has no solution since $a \in \mathbb{R}$

Could anyone confirm this equation has no solution. It's supposed to have one since it was part o an exam.
Thanks.

 

[Ray Vickson]

Homework Statement

$$z^2-\bar z |z|= 1$$

Homework Equations

Complex numbers

The Attempt at a Solution

$$(a+ib)^2-(a-ib)\sqrt{a^2+b^2} = 1$$

$$a^2-b^2+2iab-a\sqrt{a^2+b^2}+ib\sqrt{a^2+b^2} = 1$$

I separate the real part from the imaginary part.

$$\left\{\begin{matrix} 2iab+ib\sqrt{a^2+b^2}=0 \\ a^2-b^2-a\sqrt{a^2+b^2}=1 \end{matrix}\right.$$

One solution of the first is $b=0$

$$\left\{\begin{matrix} 2a+\sqrt{a^2+b^2}=0, \ 3a^2=b^2 \\ a^2-b^2-a\sqrt{a^2+b^2}=1 \end{matrix}\right.$$

So I basicalliy have two solution from the first eq.
1) $b=0$
2)$3a^2=b^2$

If you plug these results in the seconds I don't get anything...
Either I get $a^2-a^2=1$ or $-a^2=1$ which has no solution since $a \in \mathbb{R}$

Could anyone confirm this equation has no solution. It's supposed to have one since it was part o an exam.
Thanks.

We can set up the equations for the real and imaginary parts as a polynomial system, then compute a Groebner basis. This will reveal *all* the solutions. Here it is done in Maple 14. First, let z = x + I*y (I = sqrt(-1) in Maple), to get x^2-y^2 + x*r -1 = 0 and 2xy-y*r=0, where y = sqrt(x^2+y^2). We can look at a Groebner basis for the polynomial system
sys={2xy - r*y, x^2-y^2 + r*x - 1, r^2-x^2-y^2}. Maple:
with(Groebner):
B:=Basis(sys,tdeg(x,y,r));

B := [y, x - r, 2 r^2 - 1]

Therefore, a system equivalent to {sys1=0,sys2=0,sys3=0} is {B1=0,B2=0,B3=0}, or {y=0, x=r, r^2 = 1/2}.

RGV

 

[Quinzio]

Hi Ray, thanks for your time and help.

I think I notice something not correct in your setup.
You set
x^2-y^2 + x*r -1 = 0
as the real part
and
2xy-y*r=0
as imaginary part,
as I did.

Notice that the original eq is $z^2-\bar z |z|-1 = 0$
so I suppose your correct setup should be:
x^2-y^2 - x*r -1 = 0
2xy + y*r=0
one sign is flipped over in each equation.I tried to understand as well
B := [y, x - r, 2 r^2 - 1]

Therefore, a system equivalent to {sys1=0,sys2=0,sys3=0} is {B1=0,B2=0,B3=0}, or {y=0, x=r, r^2 = 1/2}.

RGV
I supposed that last one is the solution. With the sign flipped over it may have solutions I suppose.
I don't have Maple nor any evoluted math software, so if you don't mind supporting me a little bit more, please feed you software with correct eq.s and see what comes out, thanks !

 

[I like Serena]

Hi Quinzio!

In your OP you seem to be missing that $\sqrt{a^2} = |a|$.
If you take that into account, you'll start finding solutions.

 

[Quinzio]

Hello there ILS, thanks for the suggestion.
$a=\frac{-1}{\sqrt2}$

is a solution

 

[I like Serena]

Hmm.

Your equation is:
$$z^2 - \bar z |z| = 1$$

Your first solution is:
$$z = a + i b = \frac {-1} {\sqrt 2}$$

Substituting I get:
$$(\frac {-1} {\sqrt 2})^2 - {\frac {-1} {\sqrt 2}} \cdot | \frac {-1} {\sqrt 2} | ~=~ \frac 1 2 - \frac {-1} {\sqrt 2} \cdot \frac {1} {\sqrt 2} ~=~ \frac 1 2 + \frac 1 2 ~=~ 1$$



Edit: Hey, you corrected your post!