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Find the fourth roots of negative one, giving your answers in trigonometric form.
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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.
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Now, we can use de Moivreβs theorem for roots to do so.
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But before we do, weβre going to need to write the number negative one in trigonometric or polar form.
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The real part of the number negative one is negative one, and its imaginary part is zero.
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So, on an Argand diagram, itβs represented by the point whose Cartesian coordinates are negative one, zero.
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The modulus of this number is the length of the line segment that joins this point to the origin.
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So, we can see that is equal to one unit.
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And therefore, π is equal to one.
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The argument is the measure of the angle that this line segment makes with the positive real axis measured in a counterclockwise direction.
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We can see that thatβs π radians.
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And so, π, the argument of the number negative one, is equal to π.
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In polar or trigonometric form then, the number one is one times cos of π plus π sin of π.
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And now, weβre ready to use de Moivreβs theorem for roots.
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This says that for a complex number of the form π cos π plus π sin π, the πth roots are π to the power of one over π times cos of π plus two ππ over π plus π sin of π plus two ππ over π.
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And π itself takes integer values from zero through to π minus one.
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Now, weβre finding the fourth roots of negative one.
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And so, applying de Moivreβs theorem to the modulus gives us a new modulus of one to the power of one-quarter, but thatβs simply one.
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And so, we applied de Moivreβs theorem for roots to the rest of our expression, using π is equal to π.
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And, of course, since weβre finding the fourth roots π is equal to four, we get one times cos of π plus two ππ over four plus π sin of π plus two ππ over four.
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But of course, we donβt really need to include the one at the front of this expression.
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Now, since π is equal to four, weβre going to choose values of π from zero to four minus one, which is three.
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So, letβs clear some space and substitute each value of π into our general form in turn.
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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.
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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.
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And that, of course, gives us an argument of three π by four.
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Then, we substitute π equals two in.
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And we get π plus two π times two, so π plus four π over four as our argument.
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And so, our third root is cos of five π by four plus π sin of five π by four.
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Finally, we substitute π equals three into the general form for the fourth roots of negative one.
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And we get cos of seven π by four plus π sin of seven π by four.
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And whilst we could leave our answer in this form, itβs much more usual to give the argument in terms of the principal argument.
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The principal argument is the unique value of the argument that lies where the left open right closed interval negative π to π.
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And to achieve an argument within this range, we simply add or subtract multiples of two π to our given argument.
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So for one of our roots, weβre going to need to subtract two π from five π by four.
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We might write two π as eight π by four.
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So, we get five π by four minus eight π by four which is negative three π by four.
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Notice this is now within the range for the principal argument.
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Similarly, we need to subtract two π from the argument seven π by four, and that gives us negative π by four.
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And so, we rewrite our expressions for π§ sub three and π§ sub four as shown.
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Now, it doesnβt really matter how we define each of our roots.
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And so, to match the answers in this question, weβre going to switch the solutions for π§ sub three and π§ sub four.
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And we found the fourth roots of negative one.
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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.