complex numbers Q

declan_lgs · 03-05-2008 07:44PM #1

This came up on a mock paper we gotta do. Q3 (c) (ii):

Let z = x + yi, where x, y are elements of R.
Find the value of x and the value of y for which:
2z + 16 all over z + 5 = 3 - i

What do we do? I tried subbing in z but it's horrid and I can't get any values. What kind of a question is this? Has it come up before?

EamonnKeane · 03-05-2008 08:05PM

You multiply both sides by (z + 5) first, to get

2z + 16 = 3z - 15 - zi - 5i

31 + 5i = z - zi

31 + 5i = x + yi - xi + y

Therefore,
y - x = 5

and
x + y = 31

2y = 36
y = 18

x = 13

z = 13 + 18i

Pete29 · 03-05-2008 08:31PM

Never mind

alan4cult · 03-05-2008 08:58PM

2z + 16

= 3 - i
z + 5

2z + 16 = (z + 5)(3 - i)
2z + 16 = 3z + 15 - zi - 5i

2x + 2yi + 16 = 3x + 3yi + 15 - xi + y - 5i
(2x + 16) + (2y)i = (3x + 15 + y) + (3y - x - 5)i

2x + 16 = 3x + 15 + y
-x + 1 = y
x = 1 - y

2y = 3y -x -5
-y = -x -5
y = x + 5
y = 1 - y + 5
2y = 6
y = 3

x = 1 - 3 = -2

z = -2 + 3i

declan_lgs · 04-05-2008 12:40PM

Dankes!
Need to watch out for that.

JSK 252 · 04-05-2008 09:10PM

The only thing you need to remember after multiplying out is to keep real with real and imaginary with imaginary.

Peleus · 05-05-2008 12:05AM

EamonnKeane wrote: »

You multiply both sides by (z + 5) first, to get

2z + 16 = 3z - 15 - zi - 5i

31 + 5i = z - zi

31 + 5i = x + yi - xi + y

Therefore,
y - x = 5

and
x + y = 31

2y = 36
y = 18

x = 13

z = 13 + 18i

that should be a +15

conbob · 05-05-2008 10:38AM

JSK 252 wrote: »

The only thing you need to remember after multiplying out is to keep real with real and imaginary with imaginary.

keepin it real :cool:

complex numbers Q

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