Sketching Solutions for Absolute Value of (w-2j) = 3

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To solve the equation |w - 2j| = 3, the approach involves separating the cases based on the value of w relative to 2j. The solutions derived are w = 3 + 2j and w = -3 + 2j, representing two points in the complex plane. The absolute value indicates the magnitude of the complex number, which relates to the distance from the point 2j. To sketch the solutions, one should plot these points on a complex plane with real and imaginary axes. The locus of points at a distance of 3 from 2j forms a circle centered at 2j with a radius of 3.
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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...
 
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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
 
salman213 said:
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?
 
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