Simplify i^13*i^29*i^6
The given problem involves the simplification of a mathematical expression that contains powers of the imaginary unit "i". The task is to simplify the product of three different powers of "i": i^13, i^29, and i^6. The simplification process typically involves using the fact that i, which stands for the square root of -1 in complex numbers, has cyclic properties where i^2 = -1, i^3 = -i, i^4 = 1, and then the powers of i repeat in a cycle. One must apply these properties to find the simplest form of the given complex number expression.
$i^{13} \cdot i^{29} \cdot i^{6}$
Answer
Solution:
Simplification Process:
Step 1: Express $i^{13}$ as $(i^{4})^{3} \cdot i$.
Step 1.1: Extract $i^{12}$ from $i^{13}$, resulting in $i^{12} \cdot i \cdot i^{29} \cdot i^{6}$.
Step 1.2: Rewrite $i^{12}$ as $(i^{4})^{3}$, giving $(i^{4})^{3} \cdot i \cdot i^{29} \cdot i^{6}$.
Step 2: Recognize that $i^{4} = 1$.
Step 2.1: Decompose $i^{4}$ into $(i^{2})^{2}$, leading to $((i^{2})^{2})^{3} \cdot i \cdot i^{29} \cdot i^{6}$.
Step 2.2: Substitute $i^{2}$ with $-1$, to get $((-1)^{2})^{3} \cdot i \cdot i^{29} \cdot i^{6}$.
Step 2.3: Calculate $(-1)^{2}$, resulting in $1^{3} \cdot i \cdot i^{29} \cdot i^{6}$.
Step 3: Any number raised to the power of one remains unchanged, hence $1 \cdot i \cdot i^{29} \cdot i^{6}$.
Step 4: Multiply $i$ by $1$, yielding $i \cdot i^{29} \cdot i^{6}$.
Step 5: Express $i^{29}$ as $(i^{4})^{7} \cdot i$.
Step 5.1: Isolate $i^{28}$, leading to $i \cdot (i^{28} \cdot i) \cdot i^{6}$.
Step 5.2: Rewrite $i^{28}$ as $(i^{4})^{7}$, which gives $i \cdot ((i^{4})^{7} \cdot i) \cdot i^{6}$.
Step 6: Again, recognize that $i^{4} = 1$.
Step 6.1: Decompose $i^{4}$ into $(i^{2})^{2}$, resulting in $i \cdot (((i^{2})^{2})^{7} \cdot i) \cdot i^{6}$.
Step 6.2: Substitute $i^{2}$ with $-1$, to get $i \cdot (((-1)^{2})^{7} \cdot i) \cdot i^{6}$.
Step 6.3: Calculate $(-1)^{2}$, yielding $i \cdot (1^{7} \cdot i) \cdot i^{6}$.
Step 7: Any number raised to the power of one remains unchanged, hence $i \cdot (1 \cdot i) \cdot i^{6}$.
Step 8: Multiply $i$ by $1$, resulting in $i \cdot i \cdot i^{6}$.
Step 9: Perform the multiplication of $i \cdot i$.
Step 9.1: Raise $i$ to the power of $1$, giving $i^{1} \cdot i \cdot i^{6}$.
Step 9.2: Raise another $i$ to the power of $1$, resulting in $i^{1} \cdot i^{1} \cdot i^{6}$.
Step 9.3: Apply the power rule $a^{m} \cdot a^{n} = a^{m + n}$ to combine exponents, leading to $i^{1 + 1} \cdot i^{6}$.
Step 9.4: Add the exponents $1$ and $1$, yielding $i^{2} \cdot i^{6}$.
Step 10: Recognize that $i^{2} = -1$, resulting in $-1 \cdot i^{6}$.
Step 11: Extract $i^{4}$ from $i^{6}$, giving $-1 \cdot (i^{4} \cdot i^{2})$.
Step 12: Again, recognize that $i^{4} = 1$.
Step 12.1: Decompose $i^{4}$ into $(i^{2})^{2}$, resulting in $-1 \cdot ((i^{2})^{2} \cdot i^{2})$.
Step 12.2: Substitute $i^{2}$ with $-1$, to get $-1 \cdot ((-1)^{2} \cdot i^{2})$.
Step 12.3: Calculate $(-1)^{2}$, yielding $-1 \cdot (1 \cdot i^{2})$.
Step 13: Multiply $i^{2}$ by $1$, resulting in $-1 \cdot i^{2}$.
Step 14: Recognize that $i^{2} = -1$, giving $-1 \cdot -1$.
Step 15: Multiply $-1$ by $-1$, which results in $1$.
Knowledge Notes:
The imaginary unit $i$ is defined as $i = \sqrt{-1}$.
The powers of $i$ are cyclical: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and then it repeats.
For any integer $n$, $i^{4n} = 1$ because $(i^4)^n = 1^n = 1$.
To simplify powers of $i$, one can factor out multiples of 4 and use the cyclical nature of $i$'s powers.
The power rule for exponents states that $a^{m} \cdot a^{n} = a^{m + n}$.
Any number raised to the power of 0 is 1, and any number raised to the power of 1 is the number itself.
Multiplying two negative numbers results in a positive product.
