(B) [tex]\sqrt{3} (cos(\frac{\pi }{4} )+i sin(\frac{\pi }{4} ))[/tex]
De Moivre's theorem: [tex]z=r*[cos([/tex]θ[tex])+i*sin([/tex]θ[tex])}[/tex][tex]][/tex]
Raising both sides to the [tex]n^{th}[/tex] - [tex]z^{n} =r^{n} *[cos(n*[/tex]θ[tex])+i*sin(n*[/tex]θ[tex])][/tex]
The square root operation undoes the squaring operation, so we'll be dealing with [tex]n=2[/tex].
If you were to go with each choice, then you'll find that squaring choice (B) will lead to the original expression given.
[tex]z=\sqrt{3} *[cos(\frac{\pi }{4} )+i*sin(\frac{\pi }{4} )]\\z^{2} =(\sqrt{3} )*[cos(2*\frac{\pi }{4} )+i*(2*\frac{\pi }{4} )]\\z^{2} =3*[cos(\frac{\pi }{2})+i*sin(\frac{\pi }{2} )][/tex]
Therefore, the correct answer to the question is (B) [tex]\sqrt{3} (cos(\frac{\pi }{4} )+i sin(\frac{\pi }{4} ))[/tex].
Know more about square roots here:
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