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How do you find the two square roots of -1 + irad3?

Answer 1

Cora Alexander

# +-(1/sqrt2+isqrt(3/2))=+-(1+isqrt3)/sqrt2#.

It is simple to find the required square roots using

Theorem D'Moivre.

This is an Aliter.

Suppose that, #z=x+iy=sqrt(-1+isqrt3), (x,y in RR)#.

#:. z^2=(x+iy)^2=(-1+isqrt3)#.

But, #(x+iy)^2=x^2+2ixy+i^2y^2=(x^2-y^2)+2ixy#.

#:. (x+iy)^2=-1+isqrt3#,

# rArr (x^2-y^2)+i(2xy)=-1+isqrt3.#

When we compare the real and imaginary parts, we get

#x^2-y^2=-1 and 2xy=sqrt3," so that,"#

#(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2#,

#=(-1)^2+(sqrt3)^2#.

# rArr (x^2+y^2)^2=4," giving, "#

# x^2+y^2=+2......[because, x,y in RR]#.

Solving this with #x^2-y^2=-1,# we get,

#x=+-1/sqrt2, and, y=sqrt3/(2x)=+-sqrt(3/2)#.

The required square roots are therefore provided by,

#x+iy=+-(1/sqrt2+isqrt(3/2))=+-(1+isqrt3)/sqrt2#.

#color(green)("Enjoy Maths.!")#

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Answer 2

James Allen

The two square roots of (-1 + i\sqrt{3}) are (\frac{1}{2} - \frac{\sqrt{3}}{2}i) and (-\frac{1}{2} + \frac{\sqrt{3}}{2}i).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.