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Complex solve an equation
- Thread starter Amer
- Start date
- Feb 5, 2012
- 1,621
Re: Complex solve a equation
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.
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
- Feb 13, 2012
- 1,704
Re: Complex solve a equation
$\displaystyle \sqrt{x+8} - \sqrt{x} = -2$ (1)
... and the applying the standard 'double squaring' procedure...
Kind regards
$\chi$ $\sigma$
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
$\chi$ $\sigma$