A) \[\pm (1+\sqrt{3}i)\]
B) \[\pm (1-\sqrt{3}i)\]
C) \[\pm (1+\sqrt{3}i)\]
D) None of these
Correct Answer: A
Solution :
Let \[x={{(-2+2\sqrt{3}i)}^{1/2}}\] \[=\sqrt{2}{{(-1+i\sqrt{3})}^{1/2}}\] \[=\sqrt{2}{{[r(\cos \theta +i\sin \theta )]}^{1/2}}\] ...(i) Here, \[r\cos \theta =-1\] ...(ii) \[r\sin \theta =\sqrt{3}\] ...(iii) Eq.\[{{(i)}^{2}}+\]Eq.(ii) \[{{r}^{2}}=4\Rightarrow r=\pm 2\] Eq. (iii)/Eq.(ii), \[\tan \theta =-\sqrt{3}\] \[\tan \theta =\tan \frac{2\pi }{3}\] \[\Rightarrow \] \[\theta =\frac{2\pi }{3}=120\] From Eq. (i), \[x=\sqrt{2}{{[\pm 2(\cos 120+i\sin 120)]}^{1/2}}\] \[x=\sqrt{2}.\sqrt{2}\pm \{\cos 60+i\sin 60\}\] By De-moivre's theorem \[x=\pm 2\left\{ \frac{1}{2}+i.\frac{\sqrt{3}}{2} \right\}\] \[\sqrt{(-2+2\sqrt{3}i)}=x\pm (1+i\sqrt{3})\] Alternate Method Let \[z=-2+\sqrt{3}i\] \[\Rightarrow \] \[|z|=\sqrt{4+12}=4\] Now, \[\sqrt{z}=\pm \left[ \sqrt{\frac{|z|+x}{2}}+i\sqrt{\frac{|z|-x}{2}} \right]\] \[\Rightarrow \] \[\sqrt{z}=\pm \left[ \sqrt{\frac{4-2}{2}}+i\sqrt{\frac{4+2}{2}} \right]\]\[\left[ \begin{align} & Here,\,x=-2 \\ & and\,y=2\sqrt{3} \\ \end{align} \right]\] \[\Rightarrow \] \[\sqrt{z}=\pm (1+\sqrt{3}i)\]
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