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Byte of Accounting, Inc.General JournalNeed help with 27 through.pdf

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Calculate the complex number (1-i)^11? Solution To write the algebraic form of the complex number (1-i)^11, we\'ll use the fact that (1-i)^2=1- 2i+i^2, where i^2=-1. So, (1-i)^2=1-2i-1= -2i We\'ll write (1-i)^11 as [(1-i)^10]*(1-i). The exponent 10 is a multiple of 2, so (1-i)^10=[(1-i)^2]^5. (1-i)^11 = (1-i)*[(1-i)^2]^5 But (1-i)^2 = -2i, so [(1-i)^2]^5=(-2i)^5=(-2^5)(i^5). We know that i^2=-1, so i^4=1 and i^5=i (1-i)^11= (1-i)*(-2^5)(i). (1-i)^11=(-2^5)(1+i)=-32(1+i) (1-i)^11=-32-32i.

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- Calculate the complex number (1-i)^11? Solution To write the algebraic form of the complex number (1-i)^11, we'll use the fact that (1-i)^2=1- 2i+i^2, where i^2=-1. So, (1-i)^2=1-2i-1= -2i We'll write (1-i)^11 as [(1-i)^10]*(1-i). The exponent 10 is a multiple of 2, so (1-i)^10=[(1-i)^2]^5. (1-i)^11 = (1-i)*[(1-i)^2]^5 But (1-i)^2 = -2i, so [(1-i)^2]^5=(-2i)^5=(-2^5)(i^5). We know that i^2=-1, so i^4=1 and i^5=i (1-i)^11= (1-i)*(-2^5)(i). (1-i)^11=(-2^5)(1+i)=-32(1+i) (1-i)^11=-32-32i

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