High School Complex Number Solutions for |z+1| = |z+i| and |z| = 5

[matrixone]

This is a question from a competitive entrance exam ...I just want to check whether my approach is correct as i don't have the answer keys .

here is the question :

How many complex numbers z are there such that |z+ 1| = |z+i| and |z| = 5?
(A) 0
(B) 1
(C) 2
(D) 3

My approach :
let z = x+iy
Now, using |z+ 1| = |z+i|,
|(x+1)+iy| = |x+(y+1)i|
Simplifying this, i got x=y...(1)

and since |z| = 5 , we have √(x²+y²) = 5
which means (x²+y²) = 25 ...(2)

Now, plugging (1) in (2) , we get

x² = (25/2)

therefore x can take 2 values similarly y also can take 2 values...
and since x=y in the complex number ...we have 2 solutions and hence the answer is 2

 

[BvU]

Hello m1,

I would do it exactly the same way. Does that help ?
(An alternative, graphical approach is to draw a circle in the complex plane and pick the two points that have this property)

 

[matrixone]

Hello m1,

I would do it exactly the same way. Does that help ?
(An alternative, graphical approach is to draw a circle in the complex plane and pick the two points that have this property)

Thank you sir ...but just a thought about your alternative ,
i would have to pick points in the circle centered at origin having radius 5 which are equidistant from (0,1) and (1,0) ...but doesn't that pursues the same algebraic calculations ? So its not an easier alternative in any sense right ?

 

[BvU]

to pick points in the circle centered at origin having radius 5

for which a translation +1 on the real axis ends up at the same distance from the origin as a translation +1 on the imaginary axis

And yes, it's the same thing as what you do algebraically.

 

[FactChecker]

Thank you sir ...but just a thought about your alternative ,
i would have to pick points in the circle centered at origin having radius 5 which are equidistant from (0,1) and (1,0) ...but doesn't that pursues the same algebraic calculations ? So its not an easier alternative in any sense right ?

Yes, except that the possible z solutions should be equidistant from -1 and -i. It's lucky that your answers are the same.

 

[WWGD]

Following up on FactChecker's post, the number lies on the bisector, i.e., the line bisecting the segment from -1 to i , which is the x-axis. You can see that there are two solutions.