If the locus of z∈C, such that ℜ(2z+iz−1)+ℜ(2z−iz−1)=2, is a circle of radius r and center (a,b), then r215ab is equal to:
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Step-by-Step Solution
Step 1: Simplify the given equation
We are given an equation involving the real part of complex fractions. We can use the property that for any complex number w, ℜ(w)=ℜ(w). Also, (2z+iz−1)=2z−iz−1. Therefore, the second term is the conjugate of the first term, which means both terms have the same real part.
Step 2: Express in terms of x and y
Let z=x+iy. Substitute this into the simplified equation 2ℜ(2z+iz−1)=2, which simplifies to ℜ(2z+iz−1)=1. Then, multiply the numerator and denominator by the conjugate of the denominator to isolate the real part.
Step 3: Extract the real part
After multiplying the numerator and denominator, the real part of the numerator is (x−1)(2x)+y(y)=2x2−2x+y2. The denominator is (2x)2+(2y+1)2=4x2+(2y+1)2. Setting the real part equal to 1 gives the equation of the locus.
Step 4: Rearrange into standard circle equation
Multiply both sides by the denominator and rearrange the terms to form the general equation of a circle. This involves moving all terms to one side and simplifying.
Step 5: Find center and radius
Divide the entire equation by 2 to get x2+x+y2+23y+21=0. Then, complete the square for both x and y terms to find the center (a,b) and radius r. The center is (−21,−43) and r2=165.
Step 6: Calculate the final expression
Substitute the values of a=−21, b=−43, and r2=165 into the expression r215ab and simplify to get the final answer.