Question Video: Finding the Square Roots of Complex Numbers in Polar Form Mathematics • Third Year of Secondary School

Video Transcript

Determine, in trigonometric form, the square roots of negative five minus five 𝑖 all divided by negative five plus five 𝑖 all raised to the ninth power.

In this question, we’re asked to find the square roots of a given complex number. We need to give these in trigonometric form. The easiest way to find the roots of complex numbers is to use de Moivre’s theorem for the roots of complex numbers. And we recall this tells us if 𝑍 is a complex number written in trigonometric form, so 𝑍 is 𝑟 times cos 𝜃 plus 𝑖 sin 𝜃, then the 𝑛th roots of 𝑍 are all given by the following formula. They’re equal to 𝑟 to the power of one over 𝑛 multiplied by the cos of 𝜃 plus two 𝜋𝑘 over 𝑛 plus 𝑖 sin of 𝜃 plus two 𝜋𝑘 all over 𝑛, where our values of 𝑘 are integers and they vary from zero to 𝑛 minus one.

Therefore, to use this formula to find the square roots of our complex number, we’re going to want to write this in trigonometric form. And there’re several different ways of doing this. Let’s start by simplifying our expression. We might be tempted to simplify this expression by writing the numerator and the denominator separately in trigonometric form. However, we should always check if we can simplify this expression in a different way first. We can see that both our numerator and denominator share a factor of negative five. Taking this out, we get negative five times one plus 𝑖 all divided by negative five times one minus 𝑖 all raised to the ninth power. And we can cancel the shared factor of negative five in the numerator and denominator.

Next, we can simplify this further by noticing we’re dividing two complex numbers. And one way of doing this is to multiply both the numerator and denominator by the complex conjugate of the denominator. So, inside our exponent, we need to multiply by one plus 𝑖 divided by one plus 𝑖. In the numerator, we have one plus 𝑖 all squared. We can distribute this by using the binomial theorem or by using the FOIL method. Either way, we get one plus two 𝑖 plus 𝑖 squared. In the denominator, we have one minus 𝑖 multiplied by one plus 𝑖. This is a factorization of a difference between squares, so it’s one minus 𝑖 squared. And remember, we need to raise all of this to the ninth power.

We can then simplify this even further. Remember, 𝑖 is the square root of negative one. So 𝑖 squared is equal to negative one. Substituting 𝑖 squared is negative one into this expression, the numerator becomes one plus two 𝑖 minus one, which is two 𝑖. And the denominator becomes one plus one, which is two. So we get two 𝑖 over two all raised to the ninth power. We can cancel the shared factor of two. Therefore, the complex number we’re given in the question is just equal to 𝑖 to the ninth power.

We can simplify this even further. Since 𝑖 squared is negative one, 𝑖 to the fourth power must be equal to one. And since 𝑖 to the ninth power is 𝑖 to the fourth power multiplied by 𝑖 to the fourth power multiplied by 𝑖, 𝑖 to the ninth power is just equal to 𝑖. Therefore, the complex number we’ve been asked to find the square roots of is just 𝑖. And there are several different ways of finding the square roots of this number. We’re just going to use de Moivre’s formula. And to use this formula, we’re first going to need to rewrite 𝑖 in trigonometric form. And we recall to write a complex number in trigonometric form, we just need to find its modulus and its argument.

Let’s start with the modulus. The modulus of a number is its distance from the origin in an Argand diagram. Or alternatively it’s the square root of the sum of the squares of the real and imaginary parts of our complex number. And in both cases, we know the modulus of 𝑖 is equal to one. Next, we need to find the arguments of our complex number 𝑖. Remember, this is the angle measured counterclockwise from the positive 𝑥-axis that 𝑖 makes on an Argand diagram. So we can find this value from a sketch. Remember, on an Argand diagram, the 𝑥-coordinate is the real part of our complex number and the 𝑦-coordinate is the imaginary part of our complex number. And since 𝑖 has real part zero and imaginary part one, the coordinates of 𝑖 on an Argand diagram are zero, one.

So 𝑖 lies on the positive vertical axis. This means its angle measured counterclockwise from the positive horizontal axis will be 𝜋 by two. The argument of 𝑍 or the argument of 𝑖 is equal to 𝜋 by two. Therefore, we can write our complex number 𝑖 in trigonometric form by substituting the value of 𝑟 and the value of 𝜃 into the trigonometric form. Its trigonometric form is cos of 𝜋 by two plus 𝑖 sin of 𝜋 by two.

We’re now ready to use our formula to find the square roots of this complex number 𝑍. Since we want the square roots, our value of 𝑛 is two. And remember, we found the value of 𝑟 is one and 𝜃 is 𝜋 by two. So we substitute these into the formula. This then gives us the following expression for the 𝑛th roots of our complex number. They’re given by one to the power of one over two times the cos of 𝜋 by two plus two 𝜋𝑘 over two plus 𝑖 sin of 𝜋 by two plus two 𝜋𝑘 over two. And remember, our values of 𝑘 are integers, which vary between zero and 𝑛 minus one. So 𝑘 is between zero and one.

This then gives us two square roots: one when 𝑘 is zero and one when 𝑘 is equal to one. So we’ll find our first root when we substitute 𝑘 is equal to zero into this expression and simplify. First, one to the power of one-half is just equal to one. And multiplying by one won’t change its value. Next, when 𝑘 is equal to zero, two 𝜋𝑘 is equal to zero. So we just have the cos of 𝜋 by two over two, which is the cos of 𝜋 by four. And the exact same is true in our second term. When 𝑘 is zero, we have 𝑖 sin of 𝜋 by two over two, which is 𝑖 sin of 𝜋 by four. This then gives us the first square root of our complex number.

However, we’ll get a second square root if we substitute 𝑘 is equal to one. Substituting 𝑘 is equal to one into this expression, we get cos of 𝜋 by two plus two 𝜋 all over two plus 𝑖 sin of 𝜋 by two plus two 𝜋 all over two. And we can simplify this by noticing 𝜋 over two plus two 𝜋 all over two is equal to five 𝜋 by four. And we could just leave our answer like this. However, remember, we can add and subtract integer multiples of two 𝜋 from our argument. So we can also subtract two 𝜋 from this value. This would then give us negative three 𝜋 by four. This is sometimes called the principal argument of this value.

This then gives us our final answer. The square roots in trigonometric form of this number are the cos of 𝜋 by four plus 𝑖 sin of 𝜋 by four and cos of negative three 𝜋 by four plus 𝑖 sin of negative three 𝜋 by four.