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Write a general form for the roots to the equation π§ to the πth power equals one, giving your answer in polar form.
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Remember, we say that if π§ is an πth root of unity, then it satisfies the relations π§ to the πth power equals one.
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We use De Moivreβs theorem to help us solve this equation for π§.
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And that gives us a general form for the πth roots of unity.
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Remember, De Moivreβs theorem for roots states that, for a complex number of the form π times cos π plus π sin π, the πth roots are given by 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 for values of π from zero, one, all the way up to π minus one.
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Now, our equation is π§ to the πth power equals one.
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And so weβre going to begin by writing the number one in polar form.
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The real part of one is one and the imaginary part is zero.
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And so we can represent the number one on an Argand diagram by the point whose Cartesian coordinates are one, zero.
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The modulus π of this number is the length of the line segment that joins this point to the origin.
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So thatβs clearly one.
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The argument π is the measure of the angle that this line segment makes with the positive real axis measured, of course, in a counterclockwise direction.
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We can see quite clearly that π must be equal to zero.
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The number one can therefore be written in polar or trigonometric form as one times cos of zero plus π sin of zero.
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And so the original equation in our question is as shown.
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To solve this equation and find the roots, weβre going to raise both sides to the power of one over π.
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π§ to the πth power to the power of one over π is simply π§.
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And then weβre going to apply De Moivreβs theorem for roots to the right-hand side.
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The modulus becomes one to the power of one over π.
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And then π plus two ππ over π becomes zero plus two ππ over π.
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And so our expression for π§ is shown.
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But of course, we can simplify this since one to the power of one over π is simply one.
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And zero plus two ππ over π is just two ππ over π.
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And so weβve solved the equation, finding the roots, the possible values of π§.
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The general form for these roots is cos of two ππ over π plus π sin of two ππ over π.
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And itβs important to note, of course, that π will take integer values from zero through to π minus one.