What is the answer to ((1-i)/(1+i))^{100}=a+ib ?

The answer to ((1-i)/(1+i))^{100}=a+ib is a=1,b=0

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((1-i)/(1+i))^{100}=a+ib

(\frac{1 - i}{1 + i})^{100} = a + ib

a = 1, b = 0

Solution steps

(\frac{1 - i}{1 + i})^{100} = a + ib

Simplify

(\frac{1 - i}{1 + i})^{100} : 1

1 = a + ib

Rewrite

1

in standard complex form:

1 + 0 i

1 + 0 i = a + ib

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[1 = a 0 = b]

[1 = a 0 = b] : a = 1, b = 0

a = 1, b = 0

∣ z - 2 + i ∣ = 2

z^{2} - 4 i z - 1 = 0

(2 x + y) + 6 i = 5 + (x + 4 y) i

z \cdot z + 2 z = 3 + i

2 x + 3 i y = 12 + 12 i

What is the answer to ((1-i)/(1+i))^{100}=a+ib ?
The answer to ((1-i)/(1+i))^{100}=a+ib is a=1,b=0