Simplifying Powers of Complex Number $i^{20} + i^{10}$

The question asks us to express the sum of two powers of the imaginary unit, specifically $i^{20} + i^{10}$, in the standard complex number form $c + id$, where $c$ is the real part and $d$ is the imaginary part.

Understanding Powers of $i$

The imaginary unit, denoted by $i$, has a special property related to its powers. The powers of $i$ follow a cycle of 4:

• $i^1 = i$
• $i^2 = -1$
• $i^3 = i^2 \cdot i = -1 \cdot i = -i$
• $i^4 = i^2 \cdot i^2 = (-1) \cdot (-1) = 1$
• $i^5 = i^4 \cdot i = 1 \cdot i = i$

This cycle repeats. To find the value of $i^n$ for any integer $n$, we can look at the remainder when $n$ is divided by 4.

• If the remainder is 0, $i^n = i^4 = 1$.
• If the remainder is 1, $i^n = i^1 = i$.
• If the remainder is 2, $i^n = i^2 = -1$.
• If the remainder is 3, $i^n = i^3 = -i$.

Calculating $i^{20}$

To calculate $i^{20}$, we divide the exponent 20 by 4:

$$ 20 \div 4 = 5 \text{ remainder } 0 $$

Since the remainder is 0, $i^{20}$ is equal to $i^4$ or 1.

$$ i^{20} = 1 $$

Calculating $i^{10}$

To calculate $i^{10}$, we divide the exponent 10 by 4:

$$ 10 \div 4 = 2 \text{ remainder } 2 $$

Since the remainder is 2, $i^{10}$ is equal to $i^2$. We know that $i^2 = -1$.

$$ i^{10} = -1 $$

Calculating the Sum $i^{20} + i^{10}$

Now, we add the results we found:

$$ i^{20} + i^{10} = 1 + (-1) $$

$$ i^{20} + i^{10} = 1 - 1 $$

$$ i^{20} + i^{10} = 0 $$

Expressing the Result in $c + id$ Form

The number 0 can be written in the complex form $c + id$ as $0 + i \cdot 0$. Here, the real part $c = 0$ and the imaginary part $d = 0$.

Final Answer

Therefore, the complex number $i^{20} + i^{10}$ expressed in the form $c + id$ is $0 + i \cdot 0$, which simplifies to $0$.