# Complex solve an equation

#### Amer

##### Active member
how to solve this equation in complex

$$\sqrt{x+8} + 2 = \sqrt{x}$$

Thanks

#### Sudharaka

##### Well-known member
MHB Math Helper
Re: Complex solve a equation

how to solve this equation in complex

$$\sqrt{x+8} + 2 = \sqrt{x}$$

Thanks
Hi Amer, Suppose that a complex solution exists for this equation. Let,

$\sqrt{x+8}=r_1 e^{i\theta_1}\mbox{ and }\sqrt{x}=r_2 e^{i\theta_2}$

Then by $$\sqrt{x+8} + 2 = \sqrt{x}$$ we get,

$r_1\cos\theta_1+2=r_2\cos\theta_2~~~~~~~~~~~(1)$

and,

$r_1\sin\theta_1=r_2\sin\theta_2~~~~~~~~~~~(2)$

Also,

$\sqrt{x+8}=r_1 e^{i\theta_1}\mbox{ and }\sqrt{x}=r_2 e^{i\theta_2}$

$\Rightarrow r^{2}_2 e^{2i\theta_2}+8=r^{2}_1 e^{2i\theta_1}$

$\Rightarrow r^{2}_1\sin2\theta_1=r^{2}_2\sin2\theta_2$

$\Rightarrow r^{2}_1\sin\theta_1\cos\theta_1=r^{2}_2\sin\theta_2\cos\theta_2~~~~~~~~~~~(3)$

By (2) and (3),

$r_1\cos\theta_1=r_2\cos\theta_2~~~~~~~~~~~(4)$

Therefore by (1) and (4),

$2=0$

which is a contradiction. Hence there exist no solutions for the equation,

$\sqrt{x+8} + 2 = \sqrt{x}$

Wolfram verifies this. Kind Regards,
Sudharaka.

#### chisigma

##### Well-known member
Re: Complex solve a equation

how to solve this equation in complex

$$\sqrt{x+8} + 2 = \sqrt{x}$$

Thanks
It is fully evident that x=1 [which is a complex number with imaginary part equal to zero...] is solution of the equation if one takes the negative square root in both terms. This solution can easily be found writing the equation as...

$\displaystyle \sqrt{x+8} - \sqrt{x} = -2$ (1)

... and the applying the standard 'double squaring' procedure...

Kind regards

$\chi$ $\sigma$