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Determine the solution set of the equation π§ squared equals two plus two root three π in the set of complex numbers.
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To solve this equation for π§, weβre going to need to find the square root of both sides.
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Now, on the right-hand side of our equation, we see we have a complex number.
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It has a real part of two and an imaginary part of two root three.
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To find roots of complex numbers, we know we can use De Moivreβs theorem.
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De Moivreβs theorem for roots states that we can find the πth root of a complex number of the form π cos π plus π sin π by calculating π to the power of one over π times cos of π plus two π π over π plus π sin of π plus two ππ over π, where π takes integer values from zero to π minus one.
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Now, in the case of our equation, we said weβre going to solve by finding the square root of both sides.
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So weβre going to let π be equal to two.
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This means π will be equal to zero, and it will be equal to one.
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But then we notice that the complex number in our equation is written in rectangular form, rather than polar form.
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And so letβs sketch the complex number two plus two root three π on an Argand diagram to help us convert it.
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Since the complex number has a real part of two and an imaginary part of two root three, we represent it by a point whose Cartesian coordinates are two, two root three.
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We know that π is the modulus, and thatβs the length of the line segment that joins this point to the origin.
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We can use the distance formula to find the modulus.
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That is, we find the square root of the sum of the squares of the real and imaginary parts.
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So thatβs the square root of two squared plus two root three squared.
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Two root three squared becomes four times three, which is 12.
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So we get the square root of four plus 12.
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Thatβs root 16, which is of course equal to four.
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The length of the line segment is four units, and so π, the modulus of our complex number, is four.
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Next, we want to find the argument.
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Now we know that the argument is the angle this line segment makes with the positive real axis measured in a counterclockwise direction.
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And so, for complex numbers plotted in the first quadrant, we can use the tangent ratio.
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On this occasion, the side opposite the included angle of π is two root three units, and the side adjacent to it is two units.
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We divide both the numerator and denominator of this fraction by two.
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And we get that tan π is equal to the square root of three.
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Now, in fact, we know this value.
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We know that tan of π by three radians is equal to the square root of three.
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And so the argument π is π by three.
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And so we have our complex number written in polar form.
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Itβs four cos π by three plus π sin of π by three.
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Weβre now ready to apply De Moivreβs theorem with π equals two.
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The modulus is four to the power of one-half.
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But of course, four to the power of one-half is the square root of four, so itβs just two.
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The general form of the argument to our solution is π by three plus two ππ over two.
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To find the exact solutions, weβre going to substitute π equals zero and π equals one into this expression.
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When π is equal to zero, we get two cos of π by three plus zero over two plus π sin of π by three plus zero over two.
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But of course, π by three plus zero is just π by three.
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So our numerator is π by three, and we now need to divide that by two.
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When we do, we find the first solution to our equation to be two times cos of π by six plus π sin of π by six.
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But cos of π by six is root three over two and sin of π by six is one-half, and so we can distribute these parentheses.
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Two times the square root of three over two is the square root of three.
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And two times one-half π is just π, and so we have the first solution to our equation.
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To find the second, weβre going to substitute π equals one into the general form.
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This time, we get an argument of π by three plus two π over two.
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π by three plus two π becomes seven π over three.
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And then when we divide that by two, we get seven π by six.
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And so we have two times cos of seven π by six plus π sin of seven π by six.
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We evaluate cos of seven π by six, and we get negative root three over two.
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Similarly, sin of seven π by six is negative one-half.
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Finally, we distribute those parentheses, and that gives us the second solution to our equation as negative root three minus π.
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And so weβve used De Moivreβs theorem to solve the equation.
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We write this in set notation as shown.
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And so we see that the solution set to our equation in the set of complex numbers is the set including the elements root three plus π and negative root three minus π.