Integer solutions to $|-3+p|=5$ Given that $ = −3$ is a solution to $|x+p| = 5$, find the two values of $p$, where $p \in \Bbb{Z}$.
I am familiar with complex numbers, I knew I should get acquainted for a question like this.
From my understanding; The absolute value of a complex number is the number that is obtained by taking the real part of the complex number and the number that is obtained by taking the imaginary part of the complex number.
Thus, $|x+p| = |x + i y| = \sqrt{x^2 + y^2} = 5$ (from the question)
Thus surely that would make
0 = 5 which shouldn't be possible- I think!
How do I explain this to my Professor?
 A: As others have indicated, this is not a complex analysis question.  That is because one of the problem constraints is that $p \in \Bbb{Z}$.
Therefore, focus may be restricted to those values of $p$ that are Real numbers, that (also) happen to be integers.  Therefore,  this problem actually represents the issue of how to manage a math problem that includes an absolute value.
In general, if you have a problem that contains an expression like $|f(x)|$, the simplest approach is to consider $2$ cases.  Either $f(x) \geq 0$ or $f(x) < 0.$
The following (rather inelegant) approach may be used to attack problems of this type.

$\underline{\text{Case 1} ~(-3 + p) \geq 0}$
$(-3 + p) \geq 0 \iff p \geq 3.$ 
In that event, $|-3 + p| = (+1) \times (-3 + p) = (-3 + p).$
Therefore, in this individual case, the problem becomes 
$(-3 + p) = 5 \implies p = 8.$
So, in case 1, the only solutions allowable will be values of $p$ that satisfy all of the following constraints:

*

*$p \geq 3.$

*$p = 8.$

*$p$ is an integer.

Since $p = 8$ satisfies all of the above constraints, this represents the Case 1 solution.

$\underline{\text{Case 2} ~(-3 + p) < 0}$
$(-3 + p) < 0 \iff p < 3.$ 
In that event, $|-3 + p| = (-1) \times (-3 + p) = (3 - p).$
Therefore, in this individual case, the problem becomes 
$(3 - p) = 5 \implies p = -2.$
So, in case 2, the only solutions allowable will be values of $p$ that satisfy all of the following constraints:

*

*$p < 3.$

*$p = -2.$

*$p$ is an integer.

Since $p = -2$ satisfies all of the above constraints, this represents the Case 2 solution.

Final Answer:
Case 1: $p = 8.$
Case 2: $p = -2.$
Therefore, the two solutions are $p \in \{-2,8\}.$
A: As you correctly managed to work out, you are looking for a solution to √x2+y2=5. So I would suggest drawing a circle of radius 5 and then finding the y values where x = -3. This should give you your solutions for P, which may or may not be imaginary, I haven't bothered to put any thought into trying to solve
