Question

If $\left(\frac{1-i}{1+i}\right)^{100} = c + id$, then what are the values of c and d?

Answer 1

c=1, d=0

Answer 2

To find the values of $c$ and $d$, we first need to simplify the expression $\left(\frac{1-i}{1+i}\right)^{100}$.

Step 1: Simplify the base of the expression. We will simplify the fraction $\frac{1-i}{1+i}$ by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of $1+i$ is $1-i$.

$\frac{1-i}{1+i} = \frac{1-i}{1+i} \times \frac{1-i}{1-i}$

Using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$ for the numerator and $(a+b)(a-b) = a^2 - b^2$ for the denominator:

$= \frac{(1-i)^2}{1^2 - i^2}$

Recall that $i^2 = -1$. Substitute this value:

$= \frac{1^2 - 2(1)(i) + i^2}{1 - (-1)}$ $= \frac{1 - 2i - 1}{1 + 1}$ $= \frac{-2i}{2}$ $= -i$

Step 2: Substitute the simplified base back into the original expression. Now the expression becomes:

$(-i)^{100}$

Step 3: Evaluate the power of the complex number. We can write $(-i)^{100}$ as:

$(-i)^{100} = (-1)^{100} \times (i)^{100}$

Since 100 is an even number, $(-1)^{100} = 1$. So, the expression simplifies to:

$= 1 \times i^{100}$ $= i^{100}$

To evaluate $i^{100}$, we use the cyclic property of powers of $i$: $i^1 = i$ $i^2 = -1$ $i^3 = -i$ $i^4 = 1$

The powers of $i$ repeat every 4 terms. To find $i^{100}$, we divide the exponent 100 by 4: $100 \div 4 = 25$ with a remainder of 0. When the remainder is 0, $i^n = i^4 = 1$. Therefore, $i^{100} = 1$.

Step 4: Equate the result to $c+id$. We found that $\left(\frac{1-i}{1+i}\right)^{100} = 1$. Given that $\left(\frac{1-i}{1+i}\right)^{100} = c + id$. So, we have:

$c + id = 1$

We can write $1$ in the form $a+bi$ as $1 + 0i$. Comparing the real and imaginary parts: $c = 1$ $d = 0$

The values of $c$ and $d$ are $1$ and $0$ respectively.