Powers of the Imaginary Unit

For any integer $ n \in \mathbb{Z} $, the powers of the imaginary unit i behave cyclically as shown below:

\[ \begin{array}{|l|l|l|}
\hline
\text{Column 1} & \text{Column 2} & \text{Column 3} \\
\hline
i = \sqrt{-1} & i^5 = i^4 \cdot i = i & i^{4n+1} = i \\
\hline
i^2 = -1 & i^6 = i^4 \cdot i^2 = -1 & i^{4n+2} = -1 \\
\hline
i^3 = i^2 \cdot i = -i & i^7 = i^4 \cdot i^3 = -i & i^{4n+3} = -i \\
\hline
i^4 = i^2 \cdot i^2 = 1 & i^8 = i^4 \cdot i^4 = 1 & i^{4n+4} = 1 \\
\hline
\end{array} \]

\[ \vdots \]
\[ i^{4n} = (i^4)^n = 1 \]

Therefore, we can establish that:

\[ n \equiv m \pmod{4} \Rightarrow i^n = i^m \]

When finding any integer power of i, the exponent can be replaced by its remainder when divided by 4.

Examples:

Evaluate $ i^{74} $:
\[ 74 \equiv 2 \pmod{4} \Rightarrow i^{74} = i^2 = -1 \]

Evaluate $ i^{1996} $:
\[ 1996 \equiv 0 \pmod{4} \Rightarrow i^{1996} = i^0 = 1 \] (Recall that any non-zero number raised to the power of 0 equals 1).

Evaluate $ i^{-17} $:
\[ -17 \equiv 3 \pmod{4} \Rightarrow i^{-17} = i^3 = -i \]

Evaluate $ i^{-103} $:
\[ -103 \equiv 1 \pmod{4} \Rightarrow i^{-103} = i^1 = i \]

Example:

Find one of the possible values of $x$ that satisfies the equation $ (x + i^{19})^{99} = -i $.

Since $ 19 \equiv 3 \pmod{4} \Rightarrow i^{19} = i^3 = -i $, the equation simplifies to:

\[ (x + i^{19})^{99} = -i \Rightarrow (x – i)^{99} = -i \]

Since $ 99 \equiv 3 \pmod{4} \Rightarrow i^{99} = i^3 = -i $, we can substitute $-i$ with $i^{99}$:

\[ \Rightarrow (x – i)^{99} = i^{99} \]
\[ \Rightarrow x – i = i \Rightarrow x = 2i \]

Example:

Find the real part and the imaginary part of the complex number $ z = i + i^2 + i^3 + \dots + i^{49} + i^{50} $.

Since the sum of any four consecutive powers of $i$ is zero, we group them as follows:

\[ \begin{array}{ccccc}
i^1 = i & i^5 = i & \dots & i^{45} = i & i^{49} = i \\
i^2 = -1 & i^6 = -1 & \dots & i^{46} = -1 & + \, i^{50} = -1 \\
i^3 = -i & i^7 = -i & \dots & i^{47} = -i & \\
+ \, i^4 = 1 & + \, i^8 = 1 & \dots & + \, i^{48} = 1 & \\
\hline & \hline &\hline &\hline & \\
0 & 0 & \dots & 0 & -1 + i
\end{array} \]

This yields $ z = -1 + i $. Therefore, $ \text{Re}(z) = -1 $ and $ \text{Im}(z) = 1 $.

Example:

Find the imaginary part of the complex number $ z = i^{-1} + i^{-2} + i^{-3} + \dots + i^{-50} $.

Let us multiply and divide the entire expression $ z $ by $ i^{50} $:

\[ z = \displaystyle \frac{i^{50} ( i^{-1} + i^{-2} + i^{-3} + \dots + i^{-50} )}{i^{50}} \]
\[ = \displaystyle \frac{i^{49} + i^{48} + i^{47} + \dots + i^0}{i^2} \]
\[ = \displaystyle \frac{i^0 + i^1 + i^2 + i^3 + i^4 + \dots + i^{48} + i^{49}}{-1} \]

Since the sum from $i^0$ to $i^{47}$ consists of 12 full cycles of sum 0, only $i^{48}$ and $i^{49}$ remain in the numerator:

\[ z = \frac{1 + i}{-1} = -1 – i \]

Thus, we find that $ \text{Im}(z) = -1 $.

QUESTION 2

What is the imaginary part of the complex number $ z = i – i^3 + i^5 – i^7 + \dots + i^{97} – i^{99} $?

\[ A) \ 25 \quad B) \ – \ 25 \quad C) \ 50 \quad D) \ -50 \quad E) \ 0 \]

Solution:

\[ z = i – i^3 + i^5 – i^7 + \dots + i^{97} – i^{99} \]
\[ = i – (-i) + i – (-i) + \dots + i – (-i) \]
\[ = 2i + 2i + \dots + 2i \]

There are exactly 25 pairs of such terms in total, so:

\[ = 25 \cdot 2i = 50i \]

Therefore, $ \text{Im}(z) = 50 $.

\[ \textbf{Answer: C} \]

QUESTION 3

What is the imaginary part of the complex number $ z = \displaystyle \frac{i^{-98} + i^{-99}}{i^{-100} + i^{-101}} $?

\[ A) \ – \ 2 \quad B) \ – \ 1 \quad C) \ 0 \quad D) \ 1 \quad E) \ 2 \]

Solution:

Let us multiply both the numerator and the denominator by $ i^{101} $:

\[ z = \displaystyle \frac{i^{-98} + i^{-99}}{i^{-100} + i^{-101}} = \displaystyle \frac{i^{101} ( i^{-98} + i^{-99} )}{i^{101} ( i^{-100} + i^{-101} )} \]
\[ = \displaystyle \frac{i^3 + i^2}{i^1 + i^0} = \displaystyle \frac{- i \ – \ 1}{i + 1} = \frac{-(i + 1)}{i + 1} = -1 \]

Since $z = -1$, it is a purely real number. Hence, $ \text{Im}(z) = 0 $.

\[ \textbf{Answer: C} \]