If z1 and z2 both satisfy z+zˉ=2∣z−1∣arg(z1−z2)=4π, then find Im(z1+z2).
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Step-by-Step Solution
Step 1: Simplify the given equation for z
Let z=x+iy. We substitute this into the given equation z+zˉ=2∣z−1∣. The term z+zˉ 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(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.
Step 3: Identify the parabolic locus
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.
Step 4: Use the argument condition
Since both z1 and z2 lie on the parabola y2=2x−1, let z1=x1+iy1 and z2=x2+iy2. The condition arg(z1−z2)=4π means that the slope of the line segment connecting z1 and z2 is tan(4π)=1. Therefore, x1−x2y1−y2=1, which implies y1−y2=x1−x2.
Step 5: Solve for the imaginary part of the sum
Subtracting the two parabolic equations, we get y12−y22=2(x1−x2). Factoring the left side gives (y1−y2)(y1+y2)=2(x1−x2). Since y1−y2=x1−x2 and x1=x2 (otherwise y1=y2 and z1=z2), we can divide by (x1−x2) to get y1+y2=2. The imaginary part of z1+z2 is Im(z1+z2)=y1+y2.