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.
Calculate the complex number (1-i)^11SolutionTo write the alg.pdf
1. 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
![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](https://image.slidesharecdn.com/calculatethecomplexnumber1-i11solutiontowritethealg-230326031946-20658de2/85/Calculate-the-complex-number-1-i-11SolutionTo-write-the-alg-pdf-1-320.jpg)
