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Question Video: Finding a General Expression for the nth Roots of Unity Mathematics

Write a general form for the roots on 𝑧^𝑛 = 1, giving your answer in polar form.

03:13

Video Transcript

Write a general form for the roots on 𝑧 to the 𝑛th power equals one, giving your answer in polar form.

We begin by recalling that if 𝑧 is an 𝑛th root of unity, then it satisfies the relation 𝑧 to the 𝑛th power equals one. We use de Moivre’s theorem to help us solve this equation for 𝑧, and that gives us a general form for the 𝑛th roots of unity. Remember, de Moivre’s theorem for roots states that for a complex number of the form π‘Ÿ multiplied by cos πœƒ plus 𝑖 sin πœƒ, then the 𝑛th roots are given by π‘Ÿ to the power of one over 𝑛 multiplied by cos of πœƒ plus two πœ‹π‘˜ over 𝑛 plus 𝑖 sin of πœƒ plus two πœ‹π‘˜ over 𝑛 for values of π‘˜ equal to zero, one, two, and so on up to 𝑛 minus one.

Now, our equation is 𝑧 to the 𝑛th power equals one, and so we’re going to begin by writing the number one in polar form. The real part of one is one and the imaginary part is zero. And so we can represent the number one on an Argand diagram by the point whose Cartesian coordinates are one, zero. The modulus π‘Ÿ of this number is the length of the line segment that joins this point to the origin, so π‘Ÿ is equal to one. The argument πœƒ is the measure of the angle that this line segment makes with the positive real axis measured in a counterclockwise direction. And we can see quite clearly from the diagram that πœƒ must be equal to zero. The number one can therefore be written in polar or trigonometric form as one multiplied by cos of zero plus 𝑖 sin of zero. And as a result, the original equation in our question is as shown.

To solve this equation and find the roots, we are going to raise both sides to the power of one over 𝑛. 𝑧 to the 𝑛th power to the power of one over 𝑛 is simply 𝑧. And then, we are going to apply de Moivre’s theorem for roots to the right-hand side. The modulus becomes one to the power of one over 𝑛. And then πœƒ plus two πœ‹π‘˜ over 𝑛 becomes zero plus two πœ‹π‘˜ over 𝑛, and our expression for 𝑧 is as shown. We can then simplify this, since one to the power of one over 𝑛 is simply one and zero plus two πœ‹π‘˜ over 𝑛 is just two πœ‹π‘˜ over 𝑛. And so we’ve solved the equation, finding the roots of the possible values of 𝑧.

The general form of these roots is cos two πœ‹π‘˜ over 𝑛 plus 𝑖 sin two πœ‹π‘˜ over 𝑛, where π‘˜ will take integer values from zero through to 𝑛 minus one.

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