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 September 15th, 2009, 03:00 PM #1 Newbie   Joined: May 2008 Posts: 16 Thanks: 0 Lots of imaginary number problems Hey I'm having a great deal of problem solving these complex number problems. I think I understand the method to solving them but I seem to get the wrong answer regardless of what I do. I think there's a big flaw in the way i'm thinking about them atm please try to clear it up =D. 1. so the first problem is to express the fraction x + yi / x-yi into rectangular form. so I began by just basically multiplying the denominator's conjugate into both terms and got x^2+2yi-y / x^2+y so simplifying that i get (x^2-y/x^2+y) + (2y/x^2+y)i The answer sheet tells me that the answer is 1-4i...? is the sheet wrong or what did I do wrong? 2. If z=3+4i, then express z+25/z in its simplest form. I'm wondering if you could get rid of the z in the denominator by just multiplying both terms by z? so 25+z^2? If you can't do that why not? Anyways I also attempted to solve it another way but I noticed two answers were possible, neither of which were correct. 3+4i + 25/3+4i (3+4i)(3-4i) + 25(3-4i)/(3+4i)(3-4i) multiplied everything by denom conjugate 9-16i^2 + 75-100i / 9-16i^2 25 + 75-100i/25 sooo 28-4i or since I noticed that the denom was real now I decided to multiply everything by 25 getting an answer of 700 - 100i? again if you can't do this why not? anyhow the answer on the answer sheet turns out to be a dazzling 6+8i 3. Now this question is just annoying and long and I think i did it right but I don't have an answer to compare to so just point out any flaws in logic in my solution. Solve for z and express it in the form a+bi (3+2i) + z = (1-i)z (3+2i)=z - zi - z z= 3+2i/-i nowww I make a huge assumption that I don't do anything to the left side, correct me if i'm wrong and if not how come? Multiplying by the conjugate z=3i + 2i^2/i^2 z=-2 +3i 4. Now I begin to run into some big problems, as i don't know how to add fractional complex numbers. (2/3+i) + (3/2+i) I attempted solving several ways, however they're all wrong. One of the ways I tried was to solve it by just multiplying both terms with the conjugates of both denominators and I ended up with the answer (259/260) - (47/130)i not even remotely correct. another method was to simply treat it as adding normal fractions, so finding the common denom and just using the new denom conjugate to simplify. I got the answer 2-i and the answer is 9/5 + 4/5i I could post the work I did to achieve both answers however I think there isn't a point as i'm just wondering for method not calculation errors for this question. If my method is correct i'll double check the math myself. However if neither method is correct please try to explain why and why the right method works. After I reached that point in my homework I figured there wasn't much of a point in continuing until I clear somethings up so if you could answer with haste it'll be much appreciated. 5. There is one more question I'd like help on however. What are the roots for 3rdroot of -1 Basically I let x=3rdr -1 put in zero property form x^3 + 1 = 0 after this i'm stumped It can't be a difference of squares so how would you solve for all possible complex roots? obviously one of the roots is -1. Thanks for your time ^.^ September 15th, 2009, 04:00 PM #2 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 717 Re: Lots of imaginary number problems Since problem 1 is completely screwed up, I'll address this one only. You have the right idea - multiply numerator and denominator by conjugate of denominator, but you did it completely wrong. You should have (x^2 - y^2 + 2ixy)/(x^2 + y^2) September 15th, 2009, 06:53 PM #3 Newbie   Joined: May 2008 Posts: 16 Thanks: 0 Re: Lots of imaginary number problems lol wow i can't believe i didn't square the y's.. alright thanks so I suppose the answer sheet in incorrect? September 16th, 2009, 05:59 AM #4 Senior Member   Joined: Mar 2009 Posts: 318 Thanks: 0 2) Is the expression z + (25/z), as posted, or did you mean (z + 25)/z? 3) Let z = a + bi. Then the left-hand side is (3 + a) + (2 + b)i, and the right-hand side is (1 - i)(a + bi) = (2b) + (b - a)i. Equate the real and imaginary parts, and solve the resulting system of equations. 4) You posted the expression [(2/3) + i] + [(3/2) + i], but then referred to "solving". Please clarify. Do you have an expression (as posted) and you need to simplify, or is there more (including an "equals" sign and a variable) and you need to solve? 5) Have you learned about deMoivre's rule yet? September 16th, 2009, 08:50 PM   #5
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 Originally Posted by stapel 2) Is the expression z + (25/z), as posted, or did you mean (z + 25)/z? 3) Let z = a + bi. Then the left-hand side is (3 + a) + (2 + b)i, and the right-hand side is (1 - i)(a + bi) = (2b) + (b - a)i. Equate the real and imaginary parts, and solve the resulting system of equations. 4) You posted the expression [(2/3) + i] + [(3/2) + i], but then referred to "solving". Please clarify. Do you have an expression (as posted) and you need to simplify, or is there more (including an "equals" sign and a variable) and you need to solve? 5) Have you learned about deMoivre's rule yet?
Alright thanks for your help, I've managed to solve all but 5 which I still don't fully understand, as for the clarification for 2) yes z +(25/z) for 4) I meant my method for simplification and no I havn't yet should but i'll check it out... again thanks for reading all that ^.^ September 17th, 2009, 03:26 PM #6 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 717 Re: Lots of imaginary number problems For 5) x^3+1=(x+1)(x^2 -x +1). I presume you can solve for the roots of x^2 - x +1=0. Tags imaginary, lots, number, problems Search tags for this page

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