Let z = x + iy. We substitute this into the given equation z + z = 2|z-1|. The term z + z simplifies to 2x, and |z-1| becomes |(x-1) + iy|. This allows us to express the equation in terms of x and y.

From the previous step, we have 2x = 2√((x-1)² + y²). Dividing by 2 and squaring both sides gives x² = (x-1)² + y². Expanding (x-1)² and simplifying the equation will reveal the locus of z.

Expanding the right side of the equation x² = (x-1)² + y² gives x² = x² - 2x + 1 + y². The x² terms cancel out, leaving 0 = -2x + 1 + y², which can be rearranged to y² = 2x - 1. This is the equation of a parabola.

Expanding the right side of the equation x² = (x-1)² + y² gives x² = x² - 2x + 1 + y². The x² terms cancel out, leaving 0 = -2x + 1 + y², which can be rearranged to y² = 2x - 1. This is the equation of a parabola.

Since both z_1 and z_2 lie on the parabola y² = 2x - 1, let z_1 = x_1 + iy_1 and z_2 = x_2 + iy_2. The condition (z_1 - z_2) = (π)/(4) means that the slope of the line segment connecting z_1 and z_2 is tan((π)/(4)) = 1. Therefore, (y_1 - y_2)/(x_1 - x_2) = 1, which implies y_1 - y_2 = x_1 - x_2.

Since both z_1 and z_2 lie on the parabola y² = 2x - 1, let z_1 = x_1 + iy_1 and z_2 = x_2 + iy_2. The condition (z_1 - z_2) = (π)/(4) means that the slope of the line segment connecting z_1 and z_2 is tan((π)/(4)) = 1. Therefore, (y_1 - y_2)/(x_1 - x_2) = 1, which implies y_1 - y_2 = x_1 - x_2.

Subtracting the two parabolic equations, we get y_1² - y_2² = 2(x_1 - x_2). Factoring the left side gives (y_1 - y_2)(y_1 + y_2) = 2(x_1 - x_2). Since y_1 - y_2 = x_1 - x_2 and x_1 ≠ x_2 (otherwise y_1=y_2 and z_1=z_2), we can divide by (x_1 - x_2) to get y_1 + y_2 = 2. The imaginary part of z_1 + z_2 is Im(z_1 + z_2) = y_1 + y_2.

Subtracting the two parabolic equations, we get y_1² - y_2² = 2(x_1 - x_2). Factoring the left side gives (y_1 - y_2)(y_1 + y_2) = 2(x_1 - x_2). Since y_1 - y_2 = x_1 - x_2 and x_1 ≠ x_2 (otherwise y_1=y_2 and z_1=z_2), we can divide by (x_1 - x_2) to get y_1 + y_2 = 2. The imaginary part of z_1 + z_2 is Im(z_1 + z_2) = y_1 + y_2.

If $z_{1}$ and $z_{2}$ both satisfy $z+\bar{z}=2|z-1| \arg \left(z_{1}-z_{2}\right)=\frac{\pi}{4}$ , then find $\operatorname{Im}\left(z_{1}+z_{2}\right)$ .

Answer: $\operatorname{Im}(z_1+z_2) = 2$

Step-by-step solution

Step 1: Simplify the given equation for z

Let $z = x + iy$ . We substitute this into the given equation $z + \bar{z} = 2|z-1|$ . The term $z + \bar{z}$ simplifies to $2x$ , and $|z-1|$ becomes $|(x-1) + iy|$ . This allows us to express the equation in terms of $x$ and $y$ .

Step 2: Solve for the locus of z

From the previous step, we have $2x = 2\sqrt{(x-1)^2 + y^2}$ . Dividing by 2 and squaring both sides gives $x^2 = (x-1)^2 + y^2$ . Expanding $(x-1)^2$ and simplifying the equation will reveal the locus of $z$ .

Step 3: Identify the parabolic locus

Expanding the right side of the equation $x^2 = (x-1)^2 + y^2$ gives $x^2 = x^2 - 2x + 1 + y^2$ . The $x^2$ terms cancel out, leaving $0 = -2x + 1 + y^2$ , which can be rearranged to $y^2 = 2x - 1$ . This is the equation of a parabola.

Step 4: Use the argument condition

Since both $z_1$ and $z_2$ lie on the parabola $y^2 = 2x - 1$ , let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$ . The condition $\arg(z_1 - z_2) = \frac{\pi}{4}$ means that the slope of the line segment connecting $z_1$ and $z_2$ is $\tan(\frac{\pi}{4}) = 1$ . Therefore, $\frac{y_1 - y_2}{x_1 - x_2} = 1$ , which implies $y_1 - y_2 = x_1 - x_2$ .

Step 5: Solve for the imaginary part of the sum

Subtracting the two parabolic equations, we get $y_1^2 - y_2^2 = 2(x_1 - x_2)$ . Factoring the left side gives $(y_1 - y_2)(y_1 + y_2) = 2(x_1 - x_2)$ . Since $y_1 - y_2 = x_1 - x_2$ and $x_1 \neq x_2$ (otherwise $y_1=y_2$ and $z_1=z_2$ ), we can divide by $(x_1 - x_2)$ to get $y_1 + y_2 = 2$ . The imaginary part of $z_1 + z_2$ is $\operatorname{Im}(z_1 + z_2) = y_1 + y_2$ .

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If z1z_{1}z1​ and z2z_{2}z2​ both satisfy z+zˉ=2∣z−1∣arg⁡(z1−z2)=π4z+\bar{z}=2|z-1| \arg \left(z_{1}-z_{2}\right)=\frac{\pi}{4}z+zˉ=2∣z−1∣arg(z1​−z2​)=4π​, then find Im⁡(z1+z2)\operatorname{Im}\left(z_{1}+z_{2}\right)Im(z1​+z2​).