AP EAMCET · Maths · Complex Number
What is the product of all the values of \((1-i)^{\frac{2}{5}}\) is equal to
- A \(-2 i\)
- B \(2 i\)
- C \(-2\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(-2 i\)
Step-by-step Solution
Detailed explanation
Let \(z=(1-i)^{2 / 5}\) \(\begin{aligned} & z=\left[(1-i)^2\right]^{1 / 5} \\ & z=(1-1-2 i)^{1 / 5} \\ & z=(-2 i)^{1 / 5} \Rightarrow z^5=-2 i \Rightarrow z^5+2 i=0\end{aligned}\) The product of all value of \(z\) is \(-\frac{2 i}{1}=-2 i\).
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