Question Video: Finding the Square Roots of Complex Numbers Mathematics

Determine the solution set of the equation 𝑧² = 2 + 2√(3)𝑖 in β„‚.

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Video Transcript

Determine the solution set of the equation 𝑧 squared equals two plus two root three 𝑖 in the set of complex numbers.

To solve this equation for 𝑧, we’re going to need to find the square root of both sides. Now, on the right-hand side of our equation, we see we have a complex number. It has a real part of two and an imaginary part of two root three. To find roots of complex numbers, we know we can use De Moivre’s theorem. De Moivre’s theorem for roots states that we can find the 𝑛th root of a complex number of the form π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ by calculating π‘Ÿ to the power of one over 𝑛 times cos of πœƒ plus two πœ‹ π‘˜ over 𝑛 plus 𝑖 sin of πœƒ plus two πœ‹π‘˜ over 𝑛, where π‘˜ takes integer values from zero to 𝑛 minus one. Now, in the case of our equation, we said we’re going to solve by finding the square root of both sides. So we’re going to let 𝑛 be equal to two. This means π‘˜ will be equal to zero, and it will be equal to one.

But then we notice that the complex number in our equation is written in rectangular form, rather than polar form. And so let’s sketch the complex number two plus two root three 𝑖 on an Argand diagram to help us convert it. 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. We know that π‘Ÿ is the modulus, and that’s the length of the line segment that joins this point to the origin. We can use the distance formula to find the modulus. That is, we find the square root of the sum of the squares of the real and imaginary parts. So that’s the square root of two squared plus two root three squared. Two root three squared becomes four times three, which is 12. So we get the square root of four plus 12. That’s root 16, which is of course equal to four. The length of the line segment is four units, and so π‘Ÿ, the modulus of our complex number, is four.

Next, we want to find the argument. Now we know that the argument is the angle this line segment makes with the positive real axis measured in a counterclockwise direction. And so, for complex numbers plotted in the first quadrant, we can use the tangent ratio. On this occasion, the side opposite the included angle of πœƒ is two root three units, and the side adjacent to it is two units. We divide both the numerator and denominator of this fraction by two. And we get that tan πœƒ is equal to the square root of three. Now, in fact, we know this value. We know that tan of πœ‹ by three radians is equal to the square root of three. And so the argument πœƒ is πœ‹ by three. And so we have our complex number written in polar form. It’s four cos πœ‹ by three plus 𝑖 sin of πœ‹ by three.

We’re now ready to apply De Moivre’s theorem with 𝑛 equals two. The modulus is four to the power of one-half. But of course, four to the power of one-half is the square root of four, so it’s just two. The general form of the argument to our solution is πœ‹ by three plus two πœ‹π‘˜ over two. To find the exact solutions, we’re going to substitute π‘˜ equals zero and π‘˜ equals one into this expression. When π‘˜ is equal to zero, we get two cos of πœ‹ by three plus zero over two plus 𝑖 sin of πœ‹ by three plus zero over two. But of course, πœ‹ by three plus zero is just πœ‹ by three. So our numerator is πœ‹ by three, and we now need to divide that by two.

When we do, we find the first solution to our equation to be two times cos of πœ‹ by six plus 𝑖 sin of πœ‹ by six. But cos of πœ‹ by six is root three over two and sin of πœ‹ by six is one-half, and so we can distribute these parentheses. Two times the square root of three over two is the square root of three. And two times one-half 𝑖 is just 𝑖, and so we have the first solution to our equation. To find the second, we’re going to substitute π‘˜ equals one into the general form. This time, we get an argument of πœ‹ by three plus two πœ‹ over two. πœ‹ by three plus two πœ‹ becomes seven πœ‹ over three. And then when we divide that by two, we get seven πœ‹ by six. And so we have two times cos of seven πœ‹ by six plus 𝑖 sin of seven πœ‹ by six.

We evaluate cos of seven πœ‹ by six, and we get negative root three over two. Similarly, sin of seven πœ‹ by six is negative one-half. Finally, we distribute those parentheses, and that gives us the second solution to our equation as negative root three minus 𝑖. And so we’ve used De Moivre’s theorem to solve the equation. We write this in set notation as shown. 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 𝑖 and negative root three minus 𝑖.

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