July 20th, 2014, 07:10 AM  #1 
Newbie Joined: Jul 2014 From: australia Posts: 16 Thanks: 0  real numbers
Answers to attached questions: b) x = 2 c) x=3, y=2 f) x= 4/5 y= 3/5 the question is asking: Find real numbers x and y. for c) I did: (x+yi)(2i)=8+i x+yi= 8+i/2i x+yi= (8+i/2i) (2+i/2+i) x+yi= (16+8i+2i+i^2) / (4(i^2)) x+yi= 16+10i/4(1) x+yi= 16+10i/5 x+yi= 16/5 +2i this is incorrect according to the answer... does anyone know what I may have done wrong? Could anyone start me off on b) and f) because this question is sort of different to c)? Last edited by skipjack; July 22nd, 2014 at 07:09 AM. 
July 20th, 2014, 08:05 AM  #2 
Newbie Joined: Jul 2014 From: Greece Posts: 6 Thanks: 2 
I checked c) since I don't have much time at the moment. (x+yi)(2i)=8+i x+yi= 8+i/2i x+yi= (8+i/2i) (2+i/2+i) x+yi= (16+8i+2i+i^2) / (4(i^2)) Up until this point, everything you did is correct. But you should continue like this: x+yi= (161+10i) / (4(1)) x+yi= (15+10i)/ 5 x+yi= 15/5 + (10/5)i x+yi= 3 + 2i So x=3 and y=2 I'll check the rest later if no one else does Last edited by skipjack; July 22nd, 2014 at 07:05 AM. 
July 20th, 2014, 08:45 AM  #3 
Newbie Joined: Jul 2014 From: Greece Posts: 6 Thanks: 2 
Ok, I'll give you the idea on b) and f) in case you want to solve them yourself, otherwise tell me and I will solve them b) x^2+xi= 42i x^2=4 and x= 2 You solve these, the solution is the x that works for both (can't really help with that one without coming close to the answer. ) f) You can work with this one like you did with c). You will reach something like (x+1)yi= 1/5 + (3/5)i (Not sure if this will be the exact same thing) then you work as usual, real part of the left complex number equals to real part of the right complex number and imaginary part of left number, equal to imaginary part of right number. Last edited by skipjack; July 22nd, 2014 at 07:07 AM. 
July 20th, 2014, 09:39 AM  #4  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Quote:
x+yi = (8+i/2i) (2+i/2+i) should be: x+yi = [(8+i)/(2i)][(2+i)/(2+i)] ; OK? Your next line is correct: x+yi= (16+8i+2i+i^2) / (4(i^2)) BUT since i^2 = 1, then your next line should be: x+yi = (10i + 15) / 5 OK?  
July 20th, 2014, 05:52 PM  #5 
Newbie Joined: Jul 2014 From: australia Posts: 16 Thanks: 0 
For b) I did: x^2 + xi = 4  2i x=4 x=2 x= +/ √4 x= +/ 2 but since the other x is 2, this x needs to be 2. For f) I did: (2i)([x+1]yi=1+i ([x+1]yi=(1+i)/(2i) =(1+i)/(2i) * (2+i)/(2+i) =2+i+2i+i^2/4(i)^2 =1+3i/5 ([x+1]yi=1/5+3/5i I don't know what to do next with the (x+1) or whatever we are supposed to do. Is b) correct? Last edited by skipjack; July 22nd, 2014 at 07:10 AM. 
July 20th, 2014, 08:14 PM  #6  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, jessjans11! Quote:
Equate real and imaginary components. $\quad\begin{Bmatrix}x^2 \:=\:4 \\ x \:=\:\text{}2\end{Bmatrix}$ Therefore: $\:x = \text{}2$ Quote:
Expand: $\:2x  xi + 2yi + y \;=\;8+i$ $\qquad (2x + y) + (\text{}x + 2y)i \;=\;8 + i$ Equate real and imaginary components: $\qquad \begin{Bmatrix}2x + y &=& 8 \\ \text{}x + 2y &=& 1 \end{Bmatrix}$ Solve the system of equations: $\:x \,=\,3,\;y \,=\,2$ Quote:
Expand: $\:2x+22yi  xi  i  y \:=\:1+i$ $\qquad (2xy+2) + (\text{}x  2y  1)i \;=\;1+i$ Equate real and imaginary components: $\qquad \begin{Bmatrix} 2xy + 2 &=& 1 & \Rightarrow & 2xy &=& \text{}1 \\ \text{}x  2y  1 &=& 1 & \Rightarrow & x + 2y &=& \text{}2 \end{Bmatrix}$ Solve the system: $\:x \,=\, \text{}\tfrac{4}{5},\;y \,=\, \text{}\tfrac{3}{5}$  
July 21st, 2014, 10:10 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 
x^2 + xi = 4  2i Learning student goes: xi + 2i = 4  x^2 i(x + 2) = 4  x^2 i = (4  x^2) / (x + 2) Learning student: "but teacher, if x = 2, then that's division by zero?" Teacher: "hmmm...well...continue...factor numerator..." i = [(2x)(x+2)] / (x+2) i = 2  x (hmmm...think I'll square both sides...) i^2 = x^2  4x + 4 1 = x^2  4x + 4 x^2  4x + 5 = 0 What now, teacher? 
July 22nd, 2014, 01:23 AM  #8 
Banned Camp Joined: Jul 2014 From: USA Posts: 3 Thanks: 0 
If you have problem in complex numbers please take online tutoring.

July 22nd, 2014, 02:06 AM  #9 
Newbie Joined: Jul 2014 From: australia Posts: 16 Thanks: 0 
My apologies for posting a question that I had a slight problem with. After all, who was supposed to know that a 'mathsforum' was not the place to be posting questions, especially concerning math questions and discussing it with other members of the internet community.

July 22nd, 2014, 03:20 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 
Forget about eduniche8, Jess: he's evidently a spammer...


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