# find the two square roots of the complex number 1 − (2√6)i. your answers should be in the form x + iy, where x and y are exact.

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### 1 Answer

Let `(a+ib)` be the square root of the complex number `(1-2sqrt6i)`

Then, `(a+ib)^2=1-2sqrt6i`

`rArr (a^2-b^2)+(2ab)i=1-2sqrt6i`

Now, equating real and imaginary parts we get:

`a^2-b^2=1`

and, `2ab=-2sqrt6 `

`rArr b=-sqrt6/a`

So, `a^2-(-sqrt6/a)^2=1`

`rArr a^2-6/a^2=1`

`rArr a^4-6=a^2`

`rArr a^4-a^2-6=0`

`rArr a^4-3a^2+2a^2-6=0`

`rArr a^2(a^2-3)+2(a^2-3)=0`

`rArr (a^2-3)(a^2+2)=0`

Put each of the terms separately equal to zero.

`a^2=3`

`a=+-sqrt3`

and, `a^2=-2`

`a=sqrt(-2)`

But a is real.

So,` a=+sqrt3, -sqrt3`

When `a=+sqrt3 : b=-sqrt6/sqrt3=-sqrt2`

When `a=-sqrt3 : b=-sqrt6/-sqrt3=sqrt2`

Therefore, the roots of the complex number `(1-2sqrt6i)` are:

`(sqrt3-isqrt2)` and `(-sqrt3+isqrt2)`