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Exercises · Q6

Q.The value of 1+i1−i\dfrac{1+i}{1-i} is: (A) 11 (B) −1-1 (C) ii (D) −i-i

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Multiply top and bottom by the conjugate 1+i1+i of the denominator:

1+i1−i=(1+i)(1+i)(1−i)(1+i)=(1+i)21−i2.\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{(1+i)^{2}}{1-i^{2}}.

Numerator: (1+i)2=1+2i+i2=1+2i−1=2i(1+i)^{2}=1+2i+i^{2}=1+2i-1=2i. Denominator: 1−i2=1−(−1)=21-i^{2}=1-(-1)=2. Hence

1+i1−i=2i2=i.\frac{1+i}{1-i}=\frac{2i}{2}=i.

Why the other options are wrong: the result has zero real part and a positive unit imaginary part, so it is neither 11 (A) nor −1-1 (B), and its imaginary part is +1+1, not −1-1, ruling out −i-i (D). The correct value is ii.

Check (independent recomputation): in polar terms 1+i1+i has argument π/4\pi/4 and 1−i1-i has argument −π/4-\pi/4, both with modulus 2\sqrt2; dividing gives modulus 11 and argument π/4−(−π/4)=π/2\pi/4-(-\pi/4)=\pi/2, i.e. cos⁡π2+isin⁡π2=i\cos\tfrac\pi2+i\sin\tfrac\pi2=i — matching.

✓Final answer

(C) ii

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