quick graphing q (1 Viewer)

[shkspeare]

how do you sketch

Re(z) = iz ?

i let z = x + iy then i get some crap thing =\

 

[sneaky pete]

Re(z) = -y ?
i dunno this got me stumps

 

[Grey Council]

x = ix - y

x = -y

graph that.

 

[CM_Tutor]

Grey Council, I disagree. If x = -y, then z = -y(1 + i), in which case iz isn't real.

If Re(z) = iz, then LHS is real. Thus, RHS is real. Thus, z is purely imag. The solution is the imag axis, x = 0.

 

[Grey Council]

mmm, I don't think I quite understand. :-S

In that case, wouldn't the real of z always be zero?

Besides, where did:
z = -y (1+i)
that come from?

If z = 1-i ...... (1+i doesn't lie on x=-y, does it?)
then the real of z, ie 1 is equal to the negative of the imaginary part, which is -1. negative of negative is positive, so 1 = 1.

Now I'm really confused. :-S

:-S I hate complex number.
*goes to corner to cry*


EDIT:
Take (0,10)
Re(z) = iz
0 = i (10i)
0 = -10
which isn't true.

And since subbing in any value that lies on the line x=-y works, I would have to stick by my earlier answer, x = -y.

Is it me, or did CM_Tutor show quite clearly that he is also human?

damn it, i'd better not be wrong. humph

 

[Xayma]

mmm, I don't think I quite understand. :-S

In that case, wouldn't the real of z always be zero?

No because z=x+iy
so iz=ix-y

However because it is real ix=0
ie. x=0.

 

[Grey Council]

uhuh
"real of z always be zero"
if x = 0, then real of z is zero.

???

 

[shkspeare]

crap~! thx a lot CM_Tutor & co

wot i did was squared both sides which is not allowed.. rite?? =P haha

mm Re[z] = iz
let z = x + iy
x = i(x+iy)
x = ix - y

but x is real
thus x = 0
.'. graph y = 0 [sub x = 0]
... shit or do we graph x = 0

woot.. i mean.. shit.. wots the answer

 

[Grey Council]

hang on, wtf!

x = ix - y

LHS has no imaginary part. x = -y.
:S:S:S

why is x = 0? Only reason i can think of that is cause ix has to equal zero. but if x is zero, then the real of z is zero. Taking CM_Tutor's answer, (0,5) SHOULD be a solution.
however:
re(z) = iz
0 = i(0 + 5i)
0 = -5
which is clearly not true.

I'm confused...

 

[wogboy]

The only solution to Re(z) = iz is z = 0.

Since Re(z) must be real itself, iz must be real, so z must be purely imaginary. Thus Re(z) = 0 -> iz = 0 -> z = 0.

 

[Grey Council]

ahah, thought so.

but how would you graph that? just a dot on zero?

 

[BlackJack]

mmm... yes, just the dot on the origin.
An alternate, mathematical method:
let z= x+iy
Re(z) = iz
=> Re(x+iy) = i(x+iy)
=> x (+0i) = -y + ix
You obtain two conditions: the imaginary parts are equal and the real parts are equal.
=> x = -y, 0i = xi.
=> x=0, y = 0

 

[shkspeare]

huh wtf now we've got heaps of diff solutions

just as soon as i thought it was the imaginary axis...

ppl say its just a point on the origin.... ERRR ??!?!

 

[Grey Council]

Blackjack, thats what I said. :-D only I was more confused, heheh

It isn't a locus. Its a point, (0,0)

 

[shkspeare]

oh man im confused aiyo ..

newaiz heres another q i need help with

What is the complex locus equation for the ellipse with foci (-1,0) (2,0) which passses through (2,4i)

ok i can get the cartesian one out.. but how to put it into the complex locus form for an ellipse
i.e : | |z-a| + |z-b| | = 2a

also :

1. Sketch Re( z(zbar + 2) ) = 3 ...

2. It is given that z1 + z3 - z2 - z4 = 0 and z1 - z4 = 2i(z1-z2)

Describe this quadrilateral

 

[ngai]

oh man im confused aiyo ..

newaiz heres another q i need help with

What is the complex locus equation for the ellipse with foci (-1,0) (2,0) which passses through (2,4i)

ok i can get the cartesian one out.. but how to put it into the complex locus form for an ellipse
i.e : | |z-a| + |z-b| | = 2a

also :

1. Sketch Re( z(zbar + 2) ) = 3 ...

2. It is given that z1 + z3 - z2 - z4 = 0 and z1 - z4 = 2i(z1-z2)

Describe this quadrilateral

ellipse with foci (-1,0), (2,0), thru (2,4) has equation:
(x-1/2)^2/a^2 + y^2/b^2 = 1
sub in (2,4), solve for a^2 and b^2
and u get a^2 = 81/4, b^2 = 18
so a = 9/2
so sum of distances of z from (-1,0) and (2,0) is 2a = 9
|z+1| + |z-2| = 9

1. Re(z(zbar + 2)) = 3
Re(zzbar + 2z) = 3
Re( |z|^2 + 2z) = 3
x^2 + y^2 + 2x = 3, where z = x+yi
which is a circle, u sketch that urself

2. let A,B,C,D represent z1, z2, z3, z4
z1-z2 = z4-z3, so vector BA = vector CD
so BA = CD, and BA // CD
so ABCD is a parm
but then z1-z4 = 2i(z1-z2), so AD = 2BA, and AD perpendicular to BA
so its a parm, with a right angle at A, and AD = 2AB
so a rectangle, length is twice the width

 

[ToO LaZy ^*]

quick graph question..

find the asymptote of y = (x + 1)⁴ / x⁴ + 1

 

[ngai]

y = (x + 1)4 / x4 + 1
= 1 + (4x^3 + 6x^2 + 4x) / (x^4+1)
as x approaches infinity, (4x^3 + 6x^2 + 4x) / (x^4 + 1) approach 0
so asymptote y= 1

 

[cj_bridle]

wtf? now we have heaps of questions in 1 thread... im confused...

with the first question Re(z) = iz (graph) i just used the multiplication of ordinates....

let z = x + iy taking the real part x only

then graphing z = x (the x axis)
and graphing z = i (the y axis)

then multiplying them both there is only one point which multiplies.. i.e. the origin

so i got the dot on the origin..

 

[BlackJack]

Well, yeah, you can easily arrive at the same solution using different (and correct) methods. The dot on the origin is definitely the correct answer.