Maximize real and imaginary parts

Given|1+iz|=|1−iz| and Z=a+bi implies that b=0

Proof : |1+iz|=|1−iz|
=> |1+i(a+bi)|=|1−i(a+bi)| => √[(1-b)^2 +a^2] =√ [(1+b)^2 +a^2] since |z|=√(a^2 +b^2) => (1-b)^2 =(1+b)^2 => b=0

So, z = a

replacing z=a in th inequation |z−(13+15i)|<17 we get |a−(13+15i)|<17 or, |(a−13)+15i|<17 or, √[(a-13)^2 +(15)^2] < 17 or, [(a-13)^2 +(15)^2] < (17)^2 or, (a-13)^2 < 64 or -8 < a-13 < 8 or 5 < a < 21

So the maximum possible value for a in the range is 20 Therefore the largest possible value of a+b =20

Solution 1: We first prove that if $|1+iz| = |1-iz|$ , then $z$ must be real. To see this, note that if $z=a+bi$ then $|1+iz| = |1-b+ai| = \sqrt{(1-b)^2 + a^2}$ and $|1-iz| = |1+b-ai| = \sqrt{(1+b)^2 + a^2}$ . Thus, if $|1+iz| = |1-iz|$ then $b = 0$ .

Now assuming $z = a$ is real, the inequality $|z - (13+15i)| < 17$ becomes $\sqrt{ (a-13)^2 + 15^2 } < 17$ . Since $8^2 + 15^2 = 17^2$ , we get that $(a-13)^2 < 8^2$ . Since $a$ is an integer, $a$ must be less than or equal to $20$ .

Therefore, the largest possible value of $a+b$ is $20 + 0 =$ 20.

Solution 2: Divide both sides of $|1+ iz| = |1-iz|$ by $i$ to get $| z-i| = |z+ i |$ . So $z$ must lie on the perpendicular bisector of the line segment between $i$ and $-i$ , which is the real line. Hence finding the largest value of $a+b$ is equivalent to finding the largest value of $a$ .

The condition that $|z - (13+15i)| < 17$ means that the point also lies within the circle of radius 17. Since $17^2 - 15^2 = 8^2$ , it follows that the range of $a$ satisfies $13 - 8 < a < 13 + 8$ . Since $a$ is an integer, we have $a = 20$ is the maximum.

Approaching the problem geometrically, and thinking of $iz$ as a vector, the first condition tells us that the endpoint of this vector originating from (1,0) is the same distance from the original as the opposite vector, also originating from (1,0). If $iz$ has an x-component, one of the two vectors would end up closer to the origin, so it must have no x-component. This means that $(a+bi)*i = ai - b$ must have $b = 0$ , meaning $z=a$ is a real integer.

From here, we solve the second condition algebraically, replacing $z$ with $a$ : If the magnitude of $a-(13+15i) < 17$ then $(a-13)^2 + 15^2 < 289$ , so $(a-13)^2 < 64$ . Thus, $a-13 < 8$ , or $a-13 < -8$ . The largest $a$ from this is 20, and 20+0 = 20.