Question

If |3+4i-Z|=5 then the locus of complex number Z is

Answer 1

Put z = a + ib

Then find the magnitude on left hand side..

Answer 2

Firstly you will have to combine like terms and then take the magnitude..

Answer 3

Hmm, the locus of such a format, i.e. \(|z - z_1| = a\) represents a circle.

Answer 4

Yes but I will show you how..

Answer 5

yeah, i did that! And i get a circle. But then i even think it's a hyperbola.. So, which one is correct?

Answer 6

Let us solve it and then we can say what it is..

Answer 7

Ok!

Answer 8

u kno it's a circle :|

Answer 9

On combining we get the magnitude as:

\[\sqrt{(3-a)^2 + (4-b)^2} = 5\]

Yeah it represents the Circle..

Answer 10

it IS a circle. let me show you how.
If the real-axis is the X-axis, and the imaginary axis is the Y-axis, and \(z = x+iy\), then:

\[| 3 + 4 \iota - x + \iota y\ | = 5\]
\[or, ~ \sqrt{(3-x)^2 + (4+y)^2} = 5\]

Now when you square both sides,
\[or, ~ (3-x)^2 + (4+y)^2 = 5^2\]


And this, represents a circle!

Answer 11

Where (3,4) is the centre and radius is 5..

Answer 12

Aye^

Answer 13

@apoorvk , Sth is wrong in your reply! that should be:
|3+4i-x-iy|=5

Answer 14

Yes..

Answer 15

@apoorvk centre will be(3,4) and not (3, -4)..

Answer 16

@ujjwal you are saying right..

Answer 17

Oh yeah am sorry! Please accommodate for that! thanks @waterineyes
Should be:

\[|3+4ι−x-ιy |=5\]

\[or, \sqrt{(3−x)^2+(4-y)^2}=5\]


Now when you square both sides,
\[or,~ (3−x)^2+(4-y)^2=5^2\]

Answer 18

Thanks @ujjwal as well

Answer 19

Or you can write it as:

\[(a-3)^2 + (b - 4)^2 = 5^2\]

Compare with the Standard Equation:

\[(x-h)^2 + (y-k)^2 = r^2\]

So you will get (h,k) = (3,4) and Radius = 5..

Answer 20

Shouldn't the magnitude be: \[\sqrt{(3-x)^2+i^2(4-y)}=\sqrt{(3-x)^2-(4-y)}\]

Answer 21

*(4-y)^2

Answer 22

I am concerned about - sign

Answer 23

See, if you have complex number as : a + ib

Then magnitude is given by:

\[r = \sqrt{a ^2 + b^2}\]

Answer 24

Nope - when I take the magnitude, I do not need the 'iota' thingy - because magnitude physically represents the distance from the origin in the Argand's plane, and iota is like a 'direction-provider' here.

Answer 25

i is not considered in magnitude...

Answer 26

Thanks @apoorvk and @waterineyes !! Thanks a lot!

Answer 27

Welcome dear..

Answer 28

Anytime! ;]