Simplify i^13*i^29*i^6

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:

1. The imaginary unit $i$ is defined as $i=\sqrt{-1}$.

2. The powers of $i$ are cyclical: $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, and then it repeats.

3. For any integer $n$, $i^{4n}=1$ because $(i^4{)}^n=1^n=1$.

4. To simplify powers of $i$, one can factor out multiples of 4 and use the cyclical nature of $i$'s powers.

5. The power rule for exponents states that $a^m\cdot a^n=a^{m+n}$.

6. Any number raised to the power of 0 is 1, and any number raised to the power of 1 is the number itself.

7. Multiplying two negative numbers results in a positive product.