Step 1: Expressing in exponential form.
Since \(|x| = |y| = 1\), we express them as \( x = e^{i 2\alpha} \) and \( y = e^{i 3\beta} \).
Step 2: Evaluating \( x^6 y^4 \).
\[ x^6 y^4 = e^{i(12\alpha + 12\beta)} = e^{i 12 (\alpha + \beta)} = e^{i 12 \frac{\pi}{36}} = e^{i \frac{\pi}{3}} \] \[ \frac{1}{x^6 y^4} = e^{-i \frac{\pi}{3}} \] \[ x^6 y^4 + \frac{1}{x^6 y^4} = e^{i \frac{\pi}{3}} + e^{-i \frac{\pi}{3}} = 2\cos \frac{\pi}{3} = 1 \]
Step 3: Conclusion.
Thus, \( x^6 y^4 + \frac{1}{x^6 y^4} = 1 \).
The locus of z such that \(\frac{|z-i|}{|z+i|}\)= 2, where z = x+iy. is
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