#1.) Write (1-i)^10 in a+bi form.
#2.) Write (1+i)^20 in a+bi form.
How do I do these? Thanks!
Uese De Moivre's formula.
(1-i)^10 = (√2 cis -π/4)^10 = 2^5 cis -10π/4 = 32 cis -π/2 = -32i
(1+i)^20 = (√2 cis π/4)^20 = 2^10 cis 5π = -1024
1 - i = sqrt 2 [ cos 7pi/4 + i sin 7pi/4)
e^it = cos t + i sin t
so
1-i = 2^.5 e^(7 pi/4)
(1-i)^10 = 2^5 e^(70pi/4)
= 32 e^ (70 pi/4)
how many 2 pi circles in 70 pi/4
70/8 = 9.75
so 8 full circles and 3/4 of a circle of measure 2 pi
so
= 32 e^ (3 pi/2
cos 3 pi/2 = 0
sin 3 pi/2 = -1
so
32 ( 0 -1 i
- 32 i
To write (1-i)^10 in a+bi form, you can use the binomial theorem. The formula for the binomial theorem states that (a + b)^n can be expanded as the sum of the terms:
C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n-1) * a^1 * b^(n-1) + C(n,n) * a^0 * b^n
In our case, a = 1 and b = -i, and n = 10.
To find the coefficients (C(n,k)), we use the binomial coefficient formula, which is C(n,k) = n! / (k! * (n-k)!).
Now let's expand (1-i)^10 using the above formulas.
(1-i)^10 = C(10,0) * 1^10 * (-i)^0 + C(10,1) * 1^9 * (-i)^1 + C(10,2) * 1^8 * (-i)^2 + ... + C(10,9) * 1^1 * (-i)^9 + C(10,10) * 1^0 * (-i)^10
Simplifying each term:
= 1 * 1^10 * 1^0 + 10 * 1^9 * (-i) + 45 * 1^8 * (-i)^2 + ... + 10 * 1^1 * (-i)^9 + 1 * 1^0 * (-i)^10
= 1 - 10i + 45i^2 - 120i^3 + 210i^4 - 252i^5 + 210i^6 - 120i^7 + 45i^8 - 10i^9 + i^10
Remember that i^2 = -1, i^3 = -i, i^4 = 1, and the pattern repeats every four powers of i.
Substituting these values:
= 1 - 10i + 45(-1) - 120(-i) + 210(1) - 252(-i) + 210(-1) - 120i + 45(1) - 10(-i) + i^10
= 1 - 10i - 45 + 120i + 210 - 252i - 210 - 120i + 45 - 10i + i^10
= -9 - 262i + i^10
Therefore, (1-i)^10 can be written in a+bi form as -9 - 262i.
Now let's move on to (1+i)^20.
Using the same approach, we expand (1+i)^20 using the binomial theorem.
(1+i)^20 = C(20,0) * 1^20 * (i)^0 + C(20,1) * 1^19 * (i)^1 + C(20,2) * 1^18 * (i)^2 + ... + C(20,18) * 1^1 * (i)^18 + C(20,19) * 1^0 * (i)^19 + C(20,20) * 1^0 * (i)^20
Simplifying each term:
= 1 * 1^20 * 1^0 + 20 * 1^19 * (i) + 190 * 1^18 * (i)^2 + ... + 20 * 1^1 * (i)^18 + 1 * 1^0 * (i)^20
= 1 + 20i + 190(-1) + ... + 20(-1)^18 + (i)^20
= 1 + 20i - 190 + ... + 20 + (i)^20
= -189 + 20i + i^20
Therefore, (1+i)^20 can be written in a+bi form as -189 + 20i.