SOLUTION: Find all complex numbers z such that |z - 1| = |z + 3| + |z

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Question 1209633: Find all complex numbers z such that |z - 1| = |z + 3| + |z - 2i|.

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Here's how to approach this problem:
1. **Geometric Interpretation:**
The equation |z - 1| represents the distance between the complex number z and the complex number 1 (which can be thought of as the point (1,0) in the complex plane).
|z + 3| represents the distance between z and -3 (or (-3,0)).
|z - 2i| represents the distance between z and 2i (or (0,2)).
The equation states that the distance from z to 1 is equal to the sum of the distances from z to -3 and z to 2i.
2. **Consider the Triangle Inequality:**
The triangle inequality states that for any complex numbers a, b, and c:
|a + b| ≤ |a| + |b|
Let a = z + 3 and b = z - 2i, then |z + 3 + z - 2i| ≤ |z+3| + |z-2i|
|2z + 3 - 2i| ≤ |z+3| + |z-2i|
We are given |z - 1| = |z + 3| + |z - 2i|.
If z were on the line segment joining -3 and 2i, then |z+3| + |z-2i| would equal the distance between -3 and 2i. In this case, |z-1| = |-3-2i| = √13. But z would be a real number between -3 and 0, or an imaginary number between 0 and 2i. Neither of these satisfy the equation.
3. **Try z = x + yi:**
Substitute z = x + yi into the equation:
|x + yi - 1| = |x + yi + 3| + |x + yi - 2i|
|(x - 1) + yi| = |(x + 3) + yi| + |x + (y - 2)i|
√((x - 1)² + y²) = √((x + 3)² + y²) + √(x² + (y - 2)²)
4. **Square both sides (carefully!):**
(x - 1)² + y² = (x + 3)² + y² + x² + (y - 2)² + 2√((x + 3)² + y²)(x² + (y - 2)²)
x² - 2x + 1 + y² = x² + 6x + 9 + y² + x² + y² - 4y + 4 + 2√((x + 3)² + y²)(x² + (y - 2)²)
-2x + 1 = 6x + 13 + x² + y² - 4y + 2√((x + 3)² + y²)(x² + (y - 2)²)
-8x - 12 = x² + y² - 4y + 2√((x + 3)² + y²)(x² + (y - 2)²)
This equation is getting very complicated.
5. **Consider specific cases:**
Since the geometric interpretation is difficult, and the algebraic substitution is messy, there may not be a simple solution.
If we let y=0, we have |x-1| = |x+3| + |x|.
If x > 1, x-1 = x+3+x which is impossible.
If -3< x < 1, 1-x = x+3+x which is impossible.
If x < -3, 1-x = -x-3-x so 1 = -3-x, and x=-4.
So z=-4 is one solution.
Since the algebra is so difficult, and we found one solution, it is possible there are no others.
**Conclusion:**
z = -4 is one solution. The presence of the square roots makes it very difficult to solve for other solutions algebraically. It's possible this is the only solution.

Answer by ikleyn(52786)   (Show Source):
You can put this solution on YOUR website!
.
Find all complex numbers z such that |z - 1| = |z + 3| + |z - 2i|.
~~~~~~~~~~~~~~~~~~~~~~~~~~

        In his post, @CPhill came to the conclusion that z = -4
        is the solution (one solution) to the problem.

        But easy check shows that it is not so.

Indeed, then |z-1| = |-4-1| = |-5| = 5; |z+3| = |-4+3| = |-1| = 1, |z-2i| = |-4-2i| = = = , but 5 is not equal to .

So, it disproves the statement by @CPhill that z= 4 is a solution.

It is not.

SOLUTION: Find all complex numbers z such that |z - 1| = |z + 3| + |z - 2i|.