Find the prodoct of 3 - 7t and 2 + 5t. (2) Write in the form a + bi; (3 + 1)/(10

Question

Find the prodoct of 3 - 7t and 2 + 5t. (2) Write in the form a + bi; (3 + 1)/(10 -2i). (3) For each complex number, write it in polar form. (a) 1 - 3i (b) -5 + 5t (c) 1 (4) Write in thr form a + bi; ( 3 - i)^2 (5) Write in the form a + bi; 1/(5 - 3t)^10 (6) Find all solutions of the equation x^4 - 1. (7) Find the square root(s) of 2 - 2t, (8) Find the eighth root(s) of i. IThat is, find t^1/5) (9) Let z_1 and z_2 be complex numbers that live on the unit circle. Explain why z_1 z_2 is still on the unit circle. (10) Let z = r(cos theta + i sin theta). Use geometry to explain why z^n = r^n(cos(ntheta) + i sin(ntheta))

Explanation / Answer

15.

Note that z = 0 does not satisfy the equation, so we may assume z is not 0. This allows us to divide by z^5, getting

(z + 1)^5 / z^5 = 1

or

((z + 1)/z)^5 = 1.

Noticed that when k= 0, there is no solution for equation (1), so there is only 4 roots,which make sense because (1 +z)5-z5is a polynomial of degree 4

Thus, ((z + 1)/z) must be a fifth root of unity.

Let w be some fifth root of unity. Then we have

((z + 1)/z) = w
z + 1 = zw
1 = zw - z
1 = z(w - 1)

If w = 1, then this equation is not satisfied.

If w is not 1, then we may divide by w - 1, giving

1/(w - 1) = z.

So the solutions are 1/(w - 1), where w is a fifth root of unity other than 1.

[This agrees with the fact that we expect four solutions, because simplifying (z + 1)^5 = z^5 by expanding gives 5z^4 + 10z^3 + 20z^2 + 5z + 1, a polynomial of degree 4.]

If you like, of course, you may write out explicit expressions for w: e^(2i / 5), e^(4i / 5), e^(6i / 5), e^(8i / 5).

So the four solutions are

1/(e^(2i / 5) - 1),
1/(e^(4i / 5) - 1),
1/(e^(6i / 5) - 1),
1/(e^(8i / 5) - 1).

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