Question Video: Finding the Roots of a Negative Number in Trigonometric Form of Complex Numbers | Nagwa Question Video: Finding the Roots of a Negative Number in Trigonometric Form of Complex Numbers | Nagwa

# Question Video: Finding the Roots of a Negative Number in Trigonometric Form of Complex Numbers Mathematics • Third Year of Secondary School

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Find the fourth roots of β1, giving your answers in trigonometric form.

04:48

### Video Transcript

Find the fourth roots of negative one, giving your answers in trigonometric form.

When asked to find the fourth roots of negative one, this question is essentially asking us to solve the equation π§ to the fourth power equals negative one. Now, we can use de Moivreβs theorem for roots to do so. But before we do, weβre going to need to write the number negative one in trigonometric or polar form. The real part of the number negative one is negative one, and its imaginary part is zero. So, on an Argand diagram, itβs represented by the point whose Cartesian coordinates are negative one, zero. The modulus of this number is the length of the line segment that joins this point to the origin. So, we can see that is equal to one unit. And therefore, π is equal to one.

The argument is the measure of the angle that this line segment makes with the positive real axis measured in a counterclockwise direction. We can see that thatβs π radians. And so, π, the argument of the number negative one, is equal to π. In polar or trigonometric form then, the number one is one times cos of π plus π sin of π.

And now, weβre ready to use de Moivreβs theorem for roots. This says that for a complex number of the form π cos π plus π sin π, the πth roots are 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 ππ over π. And π itself takes integer values from zero through to π minus one. Now, weβre finding the fourth roots of negative one. And so, applying de Moivreβs theorem to the modulus gives us a new modulus of one to the power of one-quarter, but thatβs simply one. And so, we applied de Moivreβs theorem for roots to the rest of our expression, using π is equal to π. And, of course, since weβre finding the fourth roots π is equal to four, we get one times cos of π plus two ππ over four plus π sin of π plus two ππ over four. But of course, we donβt really need to include the one at the front of this expression.

Now, since π is equal to four, weβre going to choose values of π from zero to four minus one, which is three. So, letβs clear some space and substitute each value of π into our general form in turn. When π is equal to zero, we get the root cos of π plus zero over four plus π sin of π plus zero over four, which simplifies to cos of π by four plus π sin of π by four. Then, when π is equal to one, we get the root cos of π plus two π over four plus π sin of π plus two π over four. And that, of course, gives us an argument of three π by four. Then, we substitute π equals two in. And we get π plus two π times two, so π plus four π over four as our argument. And so, our third root is cos of five π by four plus π sin of five π by four.

Finally, we substitute π equals three into the general form for the fourth roots of negative one. And we get cos of seven π by four plus π sin of seven π by four. And whilst we could leave our answer in this form, itβs much more usual to give the argument in terms of the principal argument. The principal argument is the unique value of the argument that lies where the left open right closed interval negative π to π. And to achieve an argument within this range, we simply add or subtract multiples of two π to our given argument.

So for one of our roots, weβre going to need to subtract two π from five π by four. We might write two π as eight π by four. So, we get five π by four minus eight π by four which is negative three π by four. Notice this is now within the range for the principal argument. Similarly, we need to subtract two π from the argument seven π by four, and that gives us negative π by four. And so, we rewrite our expressions for π§ sub three and π§ sub four as shown.

Now, it doesnβt really matter how we define each of our roots. And so, to match the answers in this question, weβre going to switch the solutions for π§ sub three and π§ sub four. And we found the fourth roots of negative one. They are cos of π by four plus π sin of π by four, cos of three π by four plus π sin of three π by four, cos of negative π by four plus π sin of negative π by four, and cos of negative three π by four plus π sin of negative three π by four.

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