Complex numbers (1 Viewer)

[mazza_728]

if z=x+iy, find x,y when

i) z/(2+i) + 25/(11i+2) = 2z/(1+i)

Can anyone show me the working to this i keep getting it wrong :angry:

 

[Harimau]

what you want to do is firstly isolate z.

2z/(1+i) - z/(2+i) = 25/(11i+2)

z[2/(1+i) - 1/(2+i)] = 25/(11i+2)

Simplifying fraction on the left, realising the denominator on the right.

z[ (2+i-1-1)/(1+i)(2+i) ] = 25(2-11i)/(2+11i)(2-11i)

z/(1+i)(2+i) = (50-275i)/(4+121)

Take the denominator of the LSH to the RHS

z = (50-275i)(1+i)(2+i)/125

z = (50-275i)(1+3i)/125

z = (50+150i-275-825i)/125

z = ( - 225 - 675i)/125

z = -9/5 - 27i/5

Let z= x+iy, and Equate Real and Imaginary Coefficients.

So x= -9/5 and y= -27/5

Okay, i did that entirely on computer and not on paper so i might have made some careless error. Anyway the process is to first isolate Z, by factorising in this case, and getting a complex number (in the form of a+ib) and equating the real and imaginary coefficients.

 

[mazza_728]

i did that and got them answers so i feel less dumb now but its not the answers in the back of the book .. they got x=2 y=-1????

 

[Harimau]

Originally posted by mazza_728
i did that and got them answers so i feel less dumb now but its not the answers in the back of the book .. they got x=2 y=-1????

Well, either the book is wrong or we both did silly mistakes What book is it from?

 

[mazza_728]

really old one i think .. all it says is revised 4 unit course, a higher school certificate course in mathematics years 11 and 12 J coroneos
???

 

[Harimau]

Originally posted by mazza_728
really old one i think .. all it says is revised 4 unit course, a higher school certificate course in mathematics years 11 and 12 J coroneos
???

Ooo James Coroneos, well i wouldn't really know since i don't have the solutions to Coroneos, so i guess we have to wait for a third part to also do this question and see what solution they get.

 

[mazza_728]

logical thankyou

 

[freaking_out]

even if the answer at the baCK is wrong- b aware of the method of this question.

 

[ae]

if z=x+iy, find x,y when

i) z/(2+i) + 25/(11i+2) = 2z/(1+i)

soln

2z/(1+i) - z/(2+i) = 25/(11i+2)

z[2/(1+i) - 1/(2+i)] = 25/(11i+2)

z[ (4+2i-1-i)/(1+i)(2+i) ] = 25(2-11i)/(2+11i)(2-11i)

z[ (3+i)/(1+3i) ] = 25(2-11i)/(2+11i)(2-11i)

z = (50-275i)(1+3i)/(3+i)(4+121)

z = (-125i+875)/125(3+i)

z = -125(i-7)/125(3+i)

z = -(i-7)/(3+i)

z = -(i-7)(3-i)/(3+i)(3-i)

z = -(10i-20) / 10

z = 2 - i

x = Re(z) = 2
y = Im (z) = -1

PS: took Harimau's soln as base and fix the mistake he did

 

[BlackJack]

z[2/(1+i) - 1/(2+i)] = 25/(11i+2)

Simplifying fraction on the left, realising the denominator on the right.

z[ (2+i-1-1)/(1+i)(2+i) ] = 25(2-11i)/(2+11i)(2-11i)

I think this line is wrong Harimau.

 

[mazza_728]

Thankyou so much ae .. everything is soo much clearer xoxo

 

[sven0023]

hmm this is how i did it (might have been done before)

z/(2+i) + 25/(11i+2) - 2z/(1+i) = 0

z [(11i+2)(1+i) - 2(11i+2)(2+i)] = - 25(2+i)(1+i)

z = -25(2+2i+i-1) / (11i+2)[1+i-2(2+i)]
z = -25(1+3i) / (11i+2)(-3-i)
z = (-25-75i) / (5-35i)
z = (-5-15i) / (1-7i)
z = (100 - 50i) / 50 {By realising}
z = 2 - i

equating re and im, x = 2 and y = -1

 

[Kirsti]

yeah, i got x=2, y=-1 with same working as sven0023
(i originally wrote y=1 when solving. that woulda cost me. oops. must be more careful)