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Question

If $$\displaystyle \left ( \frac{1+i}{1-i} \right )^3-\left ( \frac{1-i}{1+i} \right )^3=A+iB$$, then $$A, B =$$


A
0,2
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B
0,2
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C
2,0
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D
2,0
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Solution

The correct option is C $$0, -2$$
$$\displaystyle \frac{1+i}{1-i}=\frac{(1+i)^{2}}{1^{2}-i^{2}}=i$$

$$\displaystyle \frac{1-i}{1+i}=\frac{(1-i)^{2}}{1^{2}-i^{2}}=-i$$

Putting the above values obtained in the given equation

$$\displaystyle i^{3}-(-i)^{3}(=A+iB)$$

$$\displaystyle =-i-(+i)$$

$$\displaystyle =0-2i$$

$$\therefore A=0,B=-2$$

Mathematics

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