Suppose z=a+bi, where a and b are integers and i is the imaginary unit. We are given that |1+iz|=|1−iz| and |z−(13+15i)|<17. Find the largest possible value of a+b.
i is the imaginary unit, where i^2=−1.
To solve this problem, we need to understand a few concepts related to complex numbers.
Let's start with the equation |1 + iz| = |1 - iz|. The absolute value of a complex number z = a + bi is given by |z| = sqrt(a^2 + b^2). Therefore, the equation |1 + iz| = |1 - iz| can be rewritten as sqrt((1^2 + (a + 1)b)^2 + (ib)^2) = sqrt((1^2 + (a - 1)b)^2 + (-ib)^2).
Simplifying both sides of the equation, we have:
√(1 + 2ab + (a^2 + b^2)) = √(1 - 2ab + (a^2 + b^2))
Squaring both sides of the equation, we get:
1 + 2ab + (a^2 + b^2) = 1 - 2ab + (a^2 + b^2)
The terms (a^2 + b^2) cancel out, and we are left with:
4ab = 0
Since we know that a and b are integers, the only possible solution is ab = 0. This means either a = 0 or b = 0.
Now let's move on to the second part of the problem. We are given that |z - (13 + 15i)| < 17. The distance between two complex numbers z1 = a + bi and z2 = c + di is given by |z1 - z2| = sqrt((a - c)^2 + (b - d)^2). Therefore, the inequality |z - (13 + 15i)| < 17 can be rewritten as sqrt((a - 13)^2 + (b - 15)^2) < 17.
Squaring both sides of the inequality, we get:
(a - 13)^2 + (b - 15)^2 < 289
Expanding the equation, we have:
a^2 - 26a + 169 + b^2 - 30b + 225 < 289
Combining like terms, we obtain:
a^2 + b^2 - 26a - 30b + 94 < 0
Now, we need to find the largest possible value of a + b that satisfies both conditions:
1. Either a = 0 or b = 0
2. a^2 + b^2 - 26a - 30b + 94 < 0
Let's consider the cases:
Case 1: a = 0
If a = 0, then the inequality becomes b^2 - 30b + 94 < 0. Solving this quadratic inequality, we find that 5 < b < 25.
Case 2: b = 0
If b = 0, then the inequality becomes a^2 - 26a + 94 < 0. Solving this quadratic inequality, we find that 6 < a < 20.
Therefore, the largest possible value of a + b is 25 (when a = 0 and b = 25).
In conclusion, the largest possible value of a + b is 25.