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Exercise 2.1 · Q2

Q.i1948−i−1869i^{1948} - i^{-1869}

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We reduce i1948i^{1948} using the period-44 cycle, and evaluate i−1869i^{-1869} by first reducing the positive exponent 18691869 and then inverting.

Step 1. Reduce 19481948 mod 44. 1948=4×487+01948=4\times487+0, so i1948=i0=1i^{1948}=i^0=1.

Step 2. Reduce 18691869 mod 44 (to handle the negative exponent). 1869=4×467+11869=4\times467+1, so i1869=i1=ii^{1869}=i^1=i.

Step 3. Invert to get i−1869i^{-1869}. i−1869=1i1869=1i=1i⋅−i−i=−i−i2=−i1=−ii^{-1869}=\dfrac1{i^{1869}}=\dfrac1i=\dfrac1i\cdot\dfrac{-i}{-i}=\dfrac{-i}{-i^2}=\dfrac{-i}{1}=-i.

Step 4. Subtract.

i1948−i−1869=1−(−i)=1+i.i^{1948}-i^{-1869}=1-(-i)=1+i.

✓Final answer

i1948−i−1869=1+ii^{1948}-i^{-1869}=\boxed{1+i}.

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