TS EAMCET · Maths · Complex Number
If \(z=x+i y\) is a complex number such that \(\bar{z}^{\frac{1}{3}}=a+i b\), then the value of \(\frac{1}{a^2+b^2}\left(\frac{x}{a}+\frac{y}{b}\right)\) is equal to
- A -1
- B -2
- C 0
- D 2
Answer & Solution
Correct Answer
(B) -2
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