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Step-by-Step Solution
Step 1: Represent the complex numbers
Let
z_1 = x_1 + i\,y_1
and
z_2 = x_2 + i\,y_2.
We wish to find the imaginary part of
z_1 + z_2,
which is
y_1 + y_2.
Step 2: Use the given argument condition
We know
\arg\bigl(z_1 - z_2\bigr) = \tfrac{\pi}{4}.
This implies
\tan\left(\arg\left(z_1 - z_2\right)\right) = 1.
Hence,
\[
\tan^{-1}\!\Bigl(\frac{y_1 - y_2}{x_1 - x_2}\Bigr) = \frac{\pi}{4}
\quad\Longrightarrow\quad
\frac{y_1 - y_2}{x_1 - x_2} = 1
\quad\Longrightarrow\quad
y_1 - y_2 = x_1 - x_2.
\]
Step 3: Apply the condition |z - 3| = \mathrm{Re}(z) to z_1
Since |z_1 - 3| = \mathrm{Re}(z_1) = x_1, we have
\[
|z_1 - 3|^2 = x_1^2
\quad\Longrightarrow\quad
(x_1 - 3)^2 + y_1^2 = x_1^2.
\]
Expanding,
\[
x_1^2 - 6x_1 + 9 + y_1^2 = x_1^2
\quad\Longrightarrow\quad
-6x_1 + 9 + y_1^2 = 0
\quad\Longrightarrow\quad
y_1^2 = 6x_1 - 9.
\]
Step 4: Apply the same condition to z_2
Similarly, for z_2, we have
\[
|z_2 - 3| = x_2
\quad\Longrightarrow\quad
(x_2 - 3)^2 + y_2^2 = x_2^2
\quad\Longrightarrow\quad
-6x_2 + 9 + y_2^2 = 0
\quad\Longrightarrow\quad
y_2^2 = 6x_2 - 9.
\]
Step 5: Subtract the two conditions
Subtract the equations obtained for z_1 and z_2 :
\[
\bigl((x_1 - 3)^2 + y_1^2\bigr) - \bigl((x_2 - 3)^2 + y_2^2\bigr)
= x_1^2 - x_2^2.
\]
This simplifies to
\[
(x_1 - x_2)(x_1 + x_2 - 6) + (y_1 - y_2)(y_1 + y_2)
= (x_1 - x_2)(x_1 + x_2).
\]
Step 6: Factor out and use the relation y_1 - y_2 = x_1 - x_2
Notice that
(y_1 - y_2) = (x_1 - x_2).
Rewriting the above expression in terms of (x_1 - x_2) and (y_1 + y_2) allows us to simplify the equation further. After rearrangement, we get:
\[
(x_1 + x_2 - 6) + (y_1 + y_2) = (x_1 + x_2).
\]
Step 7: Solve for y_1 + y_2
From the rearranged result,
\[
y_1 + y_2
= (x_1 + x_2) - (x_1 + x_2 - 6)
= 6.
\]
Thus, we conclude
y_1 + y_2 = 6.
Step 8: Conclude the imaginary part of z_1 + z_2
The imaginary part of
z_1 + z_2 = (x_1 + x_2) + i\,(y_1 + y_2)
is
y_1 + y_2 = 6.
Therefore, the required imaginary part is
\boxed{6}.
