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Solve π§π§ star plus π§ star minus π§ equals four plus two π.
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Now, letβs just see what weβve got here.
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π§ and π§ star denote complex numbers.
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π§ itself will be in the form π plus ππ.
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π and π are real constants.
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π represents the real part of the complex number whereas π represents its imaginary part.
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π§ star is the complex conjugate.
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And thatβs sometimes represented using π bar as well.
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Now, to find the complex conjugate of a complex number, we simply change the sign of the imaginary parts.
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And so if π§, our complex number, is π plus ππ, π§ star is π minus ππ.
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And there are some interesting properties of a complex number and its conjugate.
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Firstly, when we find their product, we end up with a purely real number.
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So letβs replace π§ with π plus ππ and π§ star with π minus ππ in our complex equation.
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When we do, we find π§π§ star to be π plus ππ times π minus ππ.
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And then, we add π minus ππ and subtract π§, which is π plus ππ.
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And of course, this is all equal to four plus two π.
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Letβs distribute some of our parentheses.
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Weβll begin here by multiplying the first term in each expression.
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Thatβs π times π, which is equal to π squared.
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Weβll multiply the outer term in each expression.
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That gives us negative πππ.
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Weβll then multiply the inner terms to get πππ.
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And finally, weβll multiply the last term.
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That gives us negative ππ all squared or negative π squared π squared plus π minus ππ just remains the same.
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And then, weβre subtracting that third term, that third expression π plus ππ.
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So we get minus π minus ππ.
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Letβs see if we can neaten this up a little bit.
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We can see that π minus π is zero.
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So those cancel.
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Similarly, negative πππ plus πππ is zero.
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We then recall that π squared is equal to negative one.
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So we use this.
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And we write negative π squared π squared as negative negative one π squared.
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Similarly, we can collect like terms.
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And we have negative two ππ here.
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Simplifying a little further and we find π squared plus π squared minus two ππ is all equal to four plus two π.
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Now, this bit is really important.
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We essentially have two complex numbers.
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On the left-hand side, the real part of the complex number is simply π squared plus π squared whereas its imaginary part, remember, thatβs the coefficient of π, is negative two π.
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On the right-hand side, the real part of our complex number is four.
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And its imaginary part is two.
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And so, for the complex numbers on each side of our equation to be equal, their real parts must be equal.
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And their imaginary parts must separately be equal.
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That is, π squared plus π squared must be equal to four.
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And negative two π must be equal to two.
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Well, we can now solve that second equation.
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Weβll divide through by negative two.
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And when we do, we find that π is equal to negative one.
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Letβs substitute that into our first equation.
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When we do, we find π squared plus negative one squared equals four.
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Well, negative one squared is simply one.
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Weβll subtract one from both sides of this equation to find that π squared is equal to three.
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And then, we square root both sides, remembering that weβre going to need to take both the positive and negative square root of three.
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And we find π is either positive or negative root three.
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Now, if we go back to our original equation, we said that π§ was equal to π plus ππ.
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Well, π is either positive or negative root three.
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And π is negative one.
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So this means the two solutions to our equation are π§ equals root three minus π or π§ equals negative root three minus π.