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Given that π₯ plus π¦π is equal to four plus two π over one minus two π all to the power of a half, find all possible real values of π₯ and π¦.
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Weβre given a complex number π§ is equal to π₯ plus π¦π.
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And weβre told that this is the square root of another complex number four plus two π over one minus two π.
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The first thing weβre going to do is to simplify the complex number inside the parentheses.
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And how we do this is we multiply both the numerator and denominator of our complex number by the complex conjugate of the denominator.
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We can do this because this is actually equal to one, and anything multiplied by one is just itself.
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And we note that the denominator is one minus two π.
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Therefore, the complex conjugate is one plus two π.
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Now, multiplying out our numerators, itβs four times one, which is four, plus four times two π, which is eight π, plus two π times one, which is two π, plus two π times two π, which is four π squared.
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And in our denominator, we have one times one plus one times two π plus negative two π times one plus negative two π times positive two π.
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Thatβs negative four π squared.
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Now, remember that π is the square root of negative one, so that π squared is actually equal to negative one, so that in our numerator, four π squared is negative four.
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And in our denominator, negative four π squared is negative four times negative one, which is four.
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In our numerator, we have eight π plus two π; thatβs 10π.
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And in our denominator, two π minus two π is zero.
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So we have four plus 10π minus four over one plus four, that is, 10π over five, which is two π.
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And this means that in our parentheses, we have simply two π, so that π§ is the square root of two π.
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This in turn means that π§ squared is equal to two π.
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But remember that π§ is actually π₯ plus π¦π.
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So π§ squared is π₯ plus π¦π squared, and thatβs equal to two π.
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And distributing our parentheses on the left-hand side, we have π₯ squared plus two ππ₯π¦ plus π¦ squared times π squared is two π.
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And recalling again that π squared is negative one, this gives us π₯ squared plus two ππ₯π¦ minus π¦ squared is equal to two π.
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And making some space, we can compare now real and imaginary parts, so that π₯ squared minus π¦ squared is equal to zero since there is no real part on the right-hand side.
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And for the imaginary part, we have two π₯π¦ is equal to two.
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Now, letβs just recall that, for a complex number π§ equal to π plus ππ, the modulus of π§ is the square root of π squared plus π squared, so that the modulus of π§ all squared is π squared plus π squared.
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And we can note also that thatβs equal to the modulus of π§ squared.
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If we apply this now to our π§ squared, we have the modulus of π§ squared is equal to the modulus of two π.
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And thatβs equal to the modulus of π§ all squared, which is the modulus of π₯ plus π¦π all squared.
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And since our π₯ corresponds with π and our π¦ corresponds with π, our modulus of π§ squared is π₯ squared plus π¦ squared.
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Now, the modulus of two π is the square root of zero squared, since thereβs no real part, plus two squared.
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And thatβs the positive square root of four, which is two.
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And so we have π₯ squared plus π¦ squared is equal to two.
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And we can add this to our collection of equations.
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If we call our equation π₯ squared minus π¦ squared is equal to zero βequation oneβ and our second equation π₯ squared plus π¦ squared is equal to two βequation twoβ, we have a pair of simultaneous equations in π₯ and π¦ that we can solve.
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If we add equations one and two, we have two π₯ squared is equal to two, so that π₯ squared is equal to one.
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And taking the square root on both sides, we have π₯ equal to positive or negative one.
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So now if we go back to our first equation, equation one, which is π₯ squared minus π¦ squared is equal to zero, this tells us that π₯ squared is equal to π¦ squared.
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And if π₯ is positive or negative one, then π₯ squared is equal to one.
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And this means that π¦ squared is equal to one.
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And taking square roots on both sides, we have π¦ is positive or negative one.
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Now, if we look at our third equation, two π₯π¦ is equal to two, dividing through by two, this gives us π₯ times π¦ is equal to one.
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And what this tells us is that π₯ and π¦ must have the same sign in the same root because a positive multiplied by a positive is a positive and a negative multiplied by a negative is also a positive.
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And so we have that π₯ is positive or negative one, π¦ is positive or negative one, and they both have to be the same sign in the same root.
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So that when π₯ is positive, one π¦ is positive one.
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And when π₯ is negative one, π¦ is also negative one.
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So if π₯ plus π¦π is the complex number four plus two π over one minus two π all to the power of a half, all the possible real values of π₯ and π¦ are one, one and negative one, negative one.