### Video Transcript

Given that π§ is negative eight π,
determine the square roots of π§ without first converting to trigonometric form.

Since we want to find the square
root of the complex number π§ is negative eight π, letβs give our square root some
form and call them π€ equal to π₯ plus ππ¦ so that π₯ is the real part and π¦ is
the coefficient of the complex part of our square root. So then, if π€ is the square root
of π§, then π€ squared is π§, which is negative eight π. And since π€ is π₯ plus ππ¦, then
π₯ plus ππ¦ squared must be negative eight π. If we expand our left-hand side, we
have π₯ squared plus two ππ₯π¦ minus π¦ squared is negative eight π. And comparing real and imaginary
parts on the left-hand side, we have π₯ squared minus π¦ squared is equal to zero
and two π₯π¦ is negative eight.

Our second equation implies π₯π¦ is
negative eight over two; that is negative four. And this tells us, for one thing,
that the signs of π₯ and π¦ are not the same. So, if π₯ is positive, π¦ must be
negative and vice versa. Now, recalling that for a complex
number π§, which is π plus ππ, the modulus of π§ is the square root of π squared
plus π squared. This then means that the modulus of
π§ squared is π squared plus π squared. And applying this to our square
root π€, where π€ is π₯ plus ππ¦, we have the modulus of π€ squared is π₯ squared
plus π¦ squared. And this must be equal to the
modulus of our original complex number π§ is negative eight π. That is the square root of zero
squared plus negative eight squared, which is eight, that is, that π₯ squared plus
π¦ squared is equal to eight.

And we now have a set of two
equations we can solve for π₯ and π¦. Thatβs π₯ squared minus π¦ squared
is zero and π₯ squared plus π¦ squared is eight. Now, just tidying up and making
some room, if we call our first equation equation one, thatβs π₯ squared minus π¦
squared is zero. And equation two is π₯ squared plus
π¦ squared is eight. Adding equations one and two gives
us two π₯ squared is equal to eight. That is, π₯ squared is equal to
four so that π₯ is equal to positive or negative two.

Our equation one tells us that π₯
squared is equal to π¦ squared so that if π₯ is positive or negative two, then the
corresponding π¦-values are negative or positive two. Remembering that our square roots
were π€ which is equal to π₯ plus ππ¦ and using the values of π₯ and π¦ that we
just found, then the square roots of π§ is negative eight π are two minus two π
and negative two plus two π.