complex numbers Q

declan_lgs

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

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

Never mind

alan4cult

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

Dankes!
Need to watch out for that.

JSK 252

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

Peleus

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

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: