Question

Can somebody solve this complex numbers equation?
|z-i| |z| = |z-i|^2

Answer 1

It's modulus and not absolute value

Answer 2

i can do it the donkey way, not sure if there is a snap way to do it or not

Answer 3

Put z = x+ iy and then square both sides and you will get your answer.

Answer 4

modulus meaning the argument?

Answer 5

you can rewrite it as
\[|z-i|=|z|\] so long as \(|z-i|\neq 0\)

Answer 6

what is modulus satellite? is it the mag of the radius from center?

Answer 7

that means
\[x^2+b^2=a^2+(b-1)^2\] or
\[b^2=(b-1)^2\]

Answer 8

yeah it is the absolute value
i.e.
\[|z|=\sqrt{a^2+b^2}\] where \(z=a+bi\)

Answer 9

See here:

|x+iy - i||x+iy| = |x+iy-i|^2
=> |x+ i(y-1)||x+iy| = |x+i(y-1)|^2
squaring both sides, you get,

(x^2 + (y-1)^2)*(x^2 + y^2) = (x^2 + (y-1)^2)^2
=> (x^2 + (y-1)^2)(x^2 + (y-1)^2 - x^2 - y^2) = 0
=> (x^2 + (y-1)^2)(-2y+1)= 0
=> (x^2 + (y-1)^2) = 0 or -2y + 1 = 0

Now, because both x^2 and (y-1)^2 are more than or equal to 0.
So, for x^2 + (y-1)^2 = 0,
x^2 = 0 and (y-1)^2 = 0
=> x = 0 and y = 1

or, -2y +1 = 0
=> y = 1/2

So the required answer will be

z = i or x+(i/2).

Now verify for these values of z.

Answer 10

okk gotcha

Answer 11

so
\[|z|=|z-i|\iff \sqrt{a^2+b^2}=\sqrt{a^2+(b-1)^2}\] making
\[b^2=(b-1)^2\]

Answer 12

that gives
\[b^2=b^2-2b+1\\
-2b+1=0\\
b=\frac{1}{2}\]

Answer 13

\[|z-i||z|=|z-i|^{2}\]

Answer 14

you do it the donkey way ill try polar

Answer 15

Waiiiiit! The second member of the equation is squared!!!

Answer 16

i just did it the donkey way
i got
\[Im (z)=\frac{1}{2}\]

Answer 17

You can also solve this in this way:

|z−i||z|=|z−i|^2

=> |z-i|(|z-i| - |z|) = 0

=> |z-i| = 0 or |z-i| = |z|

=> the distance between point z and point (0,1) in argand plane is zero
or z lies on perpendicular bisector of (0,1) and (0,0)

Answer 18

Which means that z = i (from first case)
and z lies on line 2y = 1 (which is perpendicular bisector of (0,1) and (0,0))

Answer 19

This is the geometrical method and the previous one was the algebric method.

Answer 20

@vishweshshrimali5 I get it now, but when I represent it on a graph i get an intersection of two geometric figures... how do i find their coordinates?

Answer 21

one is y= 1/2 and the other is x^2+y^2=0

Answer 22

nice viswesh

Answer 23

it is not "and" it is "or"

Answer 24

Yes @satellite73 is correct it is not the intersection of curves but the union of 2 solution sets.

Answer 25

Thanks @dan815

Answer 26

if \(|z-i|=0\) i.e. if \(z=i\) then
\[|z-i||z|=0\] and
\[|z-i|^2=0\]

Answer 27

elswise \(Im(z)=\frac{1}{2}\)

Answer 28

Yup @satellite73 is exactly correct.

Answer 29

but it's modulus and not absolute value so you should put it like this:
|z-i|=0
\[\sqrt{x ^{2}+y ^{2}}\]

Answer 30

\[x^{2}+y ^{2}=0\]

Answer 31

No No

Answer 32

You should remember that its modulus of z-i so if i assume z to be x+iy, then the new complex number will become, x+i(y-1)

so you can put that

x^2 + (y-1)^2 = 0

Answer 33

ok, thanks.

Answer 34

These are the curves:



This marked point and the line both represent the solutions of the given equation.

Answer 35

Ok, thanks and where did you get from y=1?

Answer 36

Because if I put y=1/2 in the other equation I get x^2= -1/4

Answer 37

Oh that :

we got one solution as z = i => im(z) = 1

Answer 38

Sorry but I don't get it, could you explain more. And thank you for your patience

Answer 39

Just 1 min.

Answer 40

ok

Answer 41

Hi I'm back

Answer 42

Okay see the following example:

solve x(x^2-1) = 0

Answer 43

Can you tell me the required values of x

Answer 44

The values will be x = -1,0,1

Answer 45

Right ?

Answer 46

wait

Answer 47

yeah sorry I didn't see the other message

Answer 48

I got it it's okay

Answer 49

So I represent only the solutions on the plane then I get the following graph:

Answer 50

Ok. But in our case how did you come up with z=i?? What passages did you do?

Answer 51

This graph only shows the solutions but if I draw the graph of the original function,

x^3 - x = 0