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

1 Answer
Jul 27, 2018

+-(1/sqrt2+isqrt(3/2))=+-(1+isqrt3)/sqrt2±(12+i32)=±1+i32.

Explanation:

The desired square roots can be easily foung using

D'Moivre's Theorem.

Here 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.

Comparing real & imaginary parts, we have,

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).

Thus, the desired square roots are given by,

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

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