Question

HELP WITH COMPLEX NUMBERS.
*question attached below*
Will give medal and fan :)

Answer 1

Can I have some help on this problem, please?

Answer 2

any idea u have?

Answer 3

None :(

Answer 4

first put the values z=x+iy in the fraction (z+i)/(z+2)

Answer 5

then arrange the numerator and denomintor in (a+bi) form

Answer 6

for example the denominator comes a+(c+d)i then rationalize the denominator
using a-(c+d)i

Answer 7

then rearrange the result in a+bi form

Answer 8

as it is given im(z+i)/(z+2)=0,
therefore bi will be zero

Answer 9

that is the imaginary part will be zero

Answer 10

For the first part, I ended up with ](x+i(y+1)/(x+2)+iy] =0

Answer 11

so multiply the numerator and denominator with [(x+2)-iy] to rationalize the denominator

Answer 12

Give me a sec to work it out :)

Answer 13

I ended up with: x^2+2x+2iy+y^2+ix+2i+y

Answer 14

where is the denominator?

Answer 15

Oh let me do that one now :$ sorry

Answer 16

The denominator is x^2+y^2+4x+4

Answer 17

ok now arrange numerator in (a+bi) form

Answer 18

[(x^2 + y^2 +2x +y) + ((2iy+ix+2i)] ?

Answer 19

no , u have to take out 'i' common like a+(c+d)i

and it also seems wrong in calculation

Answer 20

Wouldn't a be the real part/the real values?

Answer 21

yes it would be

Answer 22

[(x^2 + y^2 +2x +y) + (i(2y+x+2)]

Answer 23

well i came up with x^2+y^2+2x+y+i(x+2y-xy+2)

Answer 24

check if u hve done it wrong

Answer 25

This is what Wolfram Alpha gave me: x^2+i (x+2 y+2)+2 x+y^2+y

Answer 26

ok sry
so now write it with denominator
in the form a+bi

Answer 27

as real part dosn't matter here so write only imaginary part=0

Answer 28

and note real part too ,it will helpful for 2nd question

Answer 29

I'm a bit confused :S

Answer 30

why

Answer 31

I don't understand why the imaginary part =0

Answer 32

lol , it is given 0 cuz u will get equation of straight line from that

Answer 33

Okay. So, we're writing the denominator in the form a+ib now?

Answer 34

no denominator doesnt have i ,u have to numerator /denominator and then arrange in (a+bi) form

Answer 35

So then, we have (x^2+i (x+2 y+2)+2 x+y^2+y)/(x^2+y^2+4x+4)

Answer 36

yes seperate it like( a/b )+(c/d)i

Answer 37

:S

Answer 38

How can I do that?

Answer 39

lol
(x^2+2 x+y^2+y)/(x^2+y^2+4x+4)+i (x+2 y+2)/(x^2+y^2+4x+4)

Answer 40

oops sry it will not equal to 1,lol

Answer 41

just put i (x+2 y+2)/(x^2+y^2+4x+4)=0 and calculate

Answer 42

as imagnery part is zero for que 1

Answer 43

When I paste this into the solver it is giving me an error

Answer 44

put in wolfram

Answer 45

Wolfram gave me the same equation I put in :O

Answer 46

u will get (x+2 y+2=0)

Answer 47

http://www.wolframalpha.com/input/?i=+i+%28x%2B2+y%2B2%29%2F%28x%5E2%2By%5E2%2B4x%2B4%29%3D0

Answer 48

see in the solution part

Answer 49

thus its a straight line proved.

Answer 50

for question 2,
(x^2+2 x+y^2+y)/(x^2+y^2+4x+4)=0

Answer 51

that is (x^2+2 x+y^2+y)=0

Answer 52

its the standard equation of circle

Answer 53

check its centre and radius

Answer 54

it will match with that given in the question

Answer 55

Yup :)

Answer 56

yo so how u feel now was that easy

Answer 57

Yup, very :D

Answer 58

One question, though..did we ignore the Im and the Re?

Answer 59

i cant get?

Answer 60

what u trying to say

Answer 61

Like in the question where it said Im(z+i/Z+2) and Re(z+i/z+2)

Answer 62

yes we only work on the data given

Answer 63

dont care about anything else

Answer 64

Alrighty. Thank you sooooo much for your time, help and patience :)

Answer 65

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