Solving Complex Number Equations
We are given two conditions for a complex number $z$: We need to find the value of $|z|^2$.
- $|z + 2| = |z - 2|$
- $\arg\left(\frac{z + 3}{z - i}\right) = \frac{\pi}{4}$
Condition 1: Modulus Equality
Let $z = x + iy$. The first condition $|z + 2| = |z - 2|$ translates to:
$ |(x+2) + iy| = |(x-2) + iy| $
Squaring both sides gives:
$ (x+2)^2 + y^2 = (x-2)^2 + y^2 $
$ x^2 + 4x + 4 + y^2 = x^2 - 4x + 4 + y^2 $
Simplifying, we get:
$ 4x = -4x $
$ 8x = 0 \implies x = 0 $
This means $z$ lies on the imaginary axis, so $z$ can be written as $z = iy$ for some real number $y$.
Condition 2: Argument
Now substitute $z = iy$ into the second condition $\arg\left(\frac{z + 3}{z - i}\right) = \frac{\pi}{4}$:
$ \arg\left(\frac{iy + 3}{iy - i}\right) = \frac{\pi}{4} $
$ \arg\left(\frac{3 + iy}{i(y-1)}\right) = \frac{\pi}{4} $
To separate the real and imaginary parts, multiply the numerator and denominator by $-i$:
$ \arg\left(\frac{(3 + iy)(-i)}{i(y-1)(-i)}\right) = \frac{\pi}{4} $
$ \arg\left(\frac{-3i - i^2 y}{-i^2 (y-1)}\right) = \frac{\pi}{4} $
$ \arg\left(\frac{y - 3i}{y-1}\right) = \frac{\pi}{4} $
$ \arg\left(\frac{y}{y-1} + \frac{-3}{y-1}i\right) = \frac{\pi}{4} $
The argument of a complex number $\frac{a+bi}{c+di}$ is related to the arctangent of the ratio of its imaginary and real parts. For the argument to be $\frac{\pi}{4}$, the real and imaginary parts must be equal:
$ \frac{y}{y-1} = \frac{-3}{y-1} $
This requires $y = -3$ (assuming $y \neq 1$).
Calculating $|z|^2$
Since $x=0$ and $y=-3$, the complex number is $z = 0 - 3i = -3i$.
Now, we calculate $|z|^2$:
$ |z|^2 = |0 - 3i|^2 = 0^2 + (-3)^2 = 0 + 9 = 9 $
Thus, $|z|^2 = 9$.
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