Expand Using the Binomial Theorem (1+i)^20

Problem

(1+i)^20

Solution

1. Identify the parameters for the Binomial Theorem (a+b)^n=(∑_k=0^n)((n/k))*a^(n−k)*b^k) where a=1 b=i and n=20

2. Apply the formula to express the expansion as a summation.

(∑_k=0^20)((20/k))*(1)^(20−k)*(i)^k)

3. Simplify the expression by noting that 1^(20−k)=1 for all k

(∑_k=0^20)((20/k))*i^k)

4. Convert to polar form to simplify the calculation of a high power, where 1+i=√(,2)*(cos(π/4)+i*sin(π/4))

5. Apply De Moivre's Theorem which states [r*(cos(θ)+i*sin(θ))]^n=r^n*(cos(n*θ)+i*sin(n*θ))

(1+i)^20=(√(,2))^20*(cos(20⋅π/4)+i*sin(20⋅π/4))

6. Evaluate the power of the modulus and the trigonometric arguments.

(√(,2))^20=(2^(1/2))^20=2^10=1024

20⋅π/4=5*π

7. Substitute the values back into the expression.

1024*(cos(5*π)+i*sin(5*π))

8. Determine the values of the trigonometric functions where cos(5*π)=−1 and sin(5*π)=0

1024*(−1+i(0))=−1024

Final Answer

(1+i)^20=−1024

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