Question

Prove that x^2 = 4 will result into + - 2 using complex numbers.

Answer 1

Answer:

a solution in which no more subtance can be dissolved at a given temperature is called saturated solution

Answer 2

$\boxed{\rm{We\:have\:x^2=4}\textrm{ and we need to show that the result is \pm 2 using complex numbers}}$

$\longrightarrow\rm{x^2=4(1)\quad\quad\quad\quad\quad\quad\quad\quad\:\:...\:Since\: 4 \:can\: be\: written\: as \:4\times 1}$

$\longrightarrow\rm{x^{2}=4(1+i 0)\quad\quad\quad\quad\quad\quad\:\:\:...\:i\:is\:an\:imaginary\:number}$

$\longrightarrow\rm{x^{2}=4(\cos 0+i \sin 0)\quad\quad\quad\:\:...\:Since\: \cos0=1 \:and \:\sin0=0}$

$\longrightarrow\rm{x=[4(\cos 0+i \sin 0)]^{1 / 2}\quad\quad...\:Taking\:Square\:Root\:on\:Both\:Sides}$

$\longrightarrow\rm{x=2[\cos (2 n \pi+0)+i \sin (2 n \pi+0)]^{1 / 2}}$

$\longrightarrow\rm{x=2[\cos (2 n \pi)+i \sin (2 n \pi)]^{1 / 2}}$

$\longrightarrow\rm{x=2\left[\cos \left(\frac{2 n \pi}{2}\right)+i \sin \left(\frac{2 n \pi}{2}\right)\right]\quad...\:(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta}$

$\longrightarrow\rm{x=2[\cos (\mathrm{n} \pi)+i \sin (\mathrm{n} \pi)]}$

$\bullet\textrm{ Put n = 0, 1}$

$\longrightarrow\rm{\textrm{ When } n=0, \Lx=2[\cos (0)+i \sin (0)]=2(0+i 0)=2}$

$\longrightarrow\rm{\textrm{ When } n=1, \:x=2[\cos (\pi)+i \sin (\pi)]=2(-1+i 0)=-2}$

$\boxed{\textrm{So using complex number roots of x ^2 = 4 are x = \pm 2 }}$