Question Video: Finding the Square Roots of Imaginary Numbers Mathematics

Determine the square roots of 𝑧, given that 𝑧 = βˆ’8𝑖.

04:15

Video Transcript

Determine the square roots of 𝑧, given that 𝑧 is equal to negative eight 𝑖.

We can use De Moivre’s theorem for roots to help us find roots of complex numbers. De Moivre’s theorem says that, for a complex number of the form π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ, the 𝑛th root is given by π‘Ÿ to the power of one over 𝑛 times cos of πœƒ plus two πœ‹π‘˜ over 𝑛 plus 𝑖 sin πœƒ plus two πœ‹π‘˜ over 𝑛, where π‘˜ takes integer values from zero to 𝑛 minus one. Now, our complex number is actually written in rectangular form. And so to be able to apply De Moivre’s theorem, we’re going to convert it into trigonometric or polar form.

The real part of this number is zero, whilst its imaginary part is negative eight. And so we can represent this on the Argand diagram as the point with Cartesian coordinates zero, negative eight. π‘Ÿ, which is the modulus of the complex number, is the length of the line segment that joins this point to the origin. And so we see that π‘Ÿ is equal to eight.

Then we find πœƒ by measuring the size of the angle from the positive real axis in a counterclockwise direction to the line segment we made. Alternatively, we can travel in the opposite direction, noting of course that a clockwise direction gives a negative value for πœƒ. This will give us a value for πœƒ for the argument that’s within the range of the principal argument. Now, actually, we see that, traveling in this direction, we travel one-quarter of a turn, so that’s πœ‹ by two radians. So our argument is negative πœ‹ by two. We therefore see that 𝑧 can be alternatively written in polar form as eight times cos of negative πœ‹ by two plus 𝑖 sin of negative πœ‹ by two.

Since we’re finding the square roots of 𝑧, we’re raising each side to the power of one-half. So let’s clear some space and perform that step. We’re going to let 𝑛 be equal to two. And so the modulus of the square root of 𝑧 will be eight to the power of one-half or the square root of eight. Then, the rest of our square root is as shown. We see that the argument is negative πœ‹ by two plus two πœ‹π‘˜ over two. And since we’re letting 𝑛 be equal to two, then π‘˜ will take values from zero to two minus one, which is one. In other words, π‘˜ can only be zero or one. So let’s see what happens when we substitute these values of π‘˜ in.

When π‘˜ is equal to zero, the square root of 𝑧 is the square root of eight times cos of negative πœ‹ by two plus zero over two plus 𝑖 sin of negative πœ‹ by two plus zero over two. Simplifying the expression inside our parentheses, and we see that the argument is equal to negative πœ‹ by four. But actually, cos of negative πœ‹ by four is root two over two, and sin of negative πœ‹ by four is negative root two over two.

And so we can represent this in rectangular form as the square root of eight times root two over two plus the square root of eight times negative root two over two 𝑖. Then the square root of eight times the square root of two is the square root of 16, which is of course equal to four. And so the square root of 16 divided by two becomes four divided by two, which is two. And so in rectangular form, the first square root of 𝑧 is two minus two 𝑖.

We’ll now let π‘˜ be equal to one. When we do, we get an argument of negative πœ‹ by two plus two πœ‹ over two. The numerator of each part simplifies to three πœ‹ by two, and then three πœ‹ by two divided by two becomes three πœ‹ by four. All that’s left now is to convert this into rectangular form too. This time, when we evaluate cos of three πœ‹ by four, we get negative root two over two, and sin of three πœ‹ by four is root two over two. And so we get the square root of eight times negative root two over two plus root eight times root two over two 𝑖, which simplifies to negative two plus two 𝑖.

In set notation, we say that the square root of 𝑧 is the set containing the elements two minus two 𝑖 and negative two plus two 𝑖.

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