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- Thread starter Amer
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- Feb 5, 2012

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Hi Amer,how to solve this equation in complex

[tex]\sqrt{x+8} + 2 = \sqrt{x} [/tex]

Thanks

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.

- Feb 13, 2012

- 1,704

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 thehow to solve this equation in complex

[tex]\sqrt{x+8} + 2 = \sqrt{x} [/tex]

Thanks

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

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

Kind regards

$\chi$ $\sigma$