Solving and sketching |w-2j| = 3 on the complex plane

[salman213]

1. Question: Sketch the solution to

absolutevalueof*(w-2j) = 3



now if i had this type of question

absolute(x-3) = 4

i would go

x-3 if x greater than 3
-(x-3) if x less than 3
and solve separately

x = 7 and x = -1



Now should i do the same approach to this question I am a little confused :(




(w-2j) if w is greater than 2j
-(w-2j) if w is less than 2j


w = 3 + 2j

and

w = -3 + 2j

are those the only two solutions ?

and if so how do i sketch them ?

should I just make an imaginary and real axis and draw two `vectors` with the above components...

 

[salman213]

or actually i just noticed that absolute value is probably the magnitude of that complex number...hmmmm

thats weird should how I approach that then :S

 

[CEL]

or actually i just noticed that absolute value is probably the magnitude of that complex number...hmmmm

thats weird should how I approach that then :S

The difference of two numbers a - b, is a vector from b to a. The absolute value is the distance. What is the locus of the points that are at a distance 3 for the point 2j?