# Question Video: Finding the Square Roots of Complex Numbers in Algebraic Form Mathematics

Given that π§ = β8π, determine the square roots of π§ without first converting to trigonometric form.

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### 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, letβs give our square root some form and call them π€ equal to π₯ plus ππ¦ 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 ππ¦, then π₯ plus ππ¦ squared must be negative eight π. If we expand our left-hand side, we have π₯ squared plus two ππ₯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 ππ, 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 ππ¦, 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¦. Thatβs 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, thatβs 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 ππ¦ 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.