1 / 8

Roots of a complex number

To view this power point, right click on the screen and choose “Full screen”. Roots of a complex number. Chapter 7 review # 31 & 35 & 33. 1 -1. #31) Find the square root of -1 + i. Change the number from a + bi form to r(cos q + i sin q ) form. Find q. Imaginary number axis.

lesa
Download Presentation

Roots of a complex number

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. To view this power point, right click on the screen and choose “Full screen”. Roots of a complex number Chapter 7 review # 31 & 35 & 33

  2. 1 -1 #31) Find the square root of -1 + i Change the number from a + bi form to r(cos q + isin q) form. Find q. Imaginary number axis Graph the ordered pair (a,b). (-1,1) q Real number axis tan q = q = arctan (-1) q = -45o = 315o in quadrant IV or q = 135o in quadrant II.

  3. r = (-1)2 + (1)2 #31) Find the square root of -1 + i Change the number from a + bi form to r(cos q + isin q) form. Since q is in quadrant II, we will use 135o . 2 -1 + i = (cos 135o + isin 135o) Now use the rule for finding the roots of a complex number. n n (cos a + isin a) r a + bi = q + 360o k . n Where a = and k = 0, 1, 2 … n – 1

  4. r = (-1)2 + (1)2 #31) Find the square root of -1 + i Change the number from a + bi form to r(cos q + isin q) form. Since q is in quadrant II, we will use 135o . 2 -1 + i = (cos 135o + isin 135o) 135o + 360o k 2 a = and k = 0 and 1 (cos 67.5o + isin 67.5o) 4 2 -1 + i = or (cos 247.5o + isin 247.5o) 4 = 2

  5. r = (-1)2 + (1)2 #31) Find the square root of -1 + i The answer could also expressed in radians. Change the number from a + bi form to r(cos q + isin q) form. 3p 4 Since q is in quadrant II, we will use . 3p 4 3p 4 2 -1 + i = (cos + i sin ) 3p 4 + 2p k a = and k = 0 and 1 2 3p 8 3p 8 (cos + i sin ) 4 2 -1 + i = or 11p 8 (cos + i sin ) 3p 8 4 = 2

  6. r = (0)2 + (1)2 We use the same method to solve an equation. # 35) Find all complex solutions of the equation. x4 – i = 0 x4 = i Find the 4th root of i. (Use 0 + i). The ordered pair is (0,1) 0 + i = 1(cos 90o + i sin 90o)

  7. Find the 4th root of i. Change the number from a + bi form to r(cos q + isin q) form. 0 + i = (cos 90o + isin 90o) 90o + 360o k 4 a = and k = 0, 1, 2, 3 4 cos 22.5o + i sin 22.5o 0 + i = = cos 112.5o + i sin 112.5o or = cos 202.5o + i sin 202.5o or = cos 292.5o + i sin 292.5o or

  8. Find the 4th root of the real number 81. #33) Change the number from a + bi form to r(cos q + isin q) form. 81 + 0i = 81(cos 0o + isin 0o) 0o + 360o k 4 a = and k = 0, 1, 2, 3 4 3(cos 0o + i sin 0o) = 3 81 = = 3(cos 90o + i sin 90o) = 3i or = 3(cos 180o + i sin 180o) = –3 or = 3(cos 270o + i sin 270o) = –3i or

More Related