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Exercise 1.1 · Q29

Q.Find the value of 3−2i(i6−i7)(1+i11)\dfrac{3-2i}{(i^6-i^7)(1+i^{11})}

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i6=−1i^6=-1, i7=i6⋅i=−ii^7=i^6\cdot i=-i, so i6−i7=−1−(−i)=−1+ii^6-i^7=-1-(-i)=-1+i. Also i11=i8+3=i3=−ii^{11}=i^{8+3}=i^3=-i, so 1+i11=1−i1+i^{11}=1-i. Denominator: (−1+i)(1−i)=−1+i+i−i2=−1+2i+1=2i(-1+i)(1-i)=-1+i+i-i^2=-1+2i+1=2i. So the expression is 3−2i2i\dfrac{3-2i}{2i}. Multiply numerator and denominator by −i-i: $\df …

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