Question

How do you simplify ${{i}^{29}}$?

Answer 1

In the above stated question we need to simplify a complex term, the complex term in the above given question is the term “i” this “i” term in complex signifies $\sqrt{-1}$ so from the above question we finally get it as we need to simplify ${{\left( \sqrt{-1} \right)}^{29}}$.

Complete step-by-step solution:
In the above mentioned question we need to simplify ${{i}^{29}}$, this term “i” which is mentioned in the question is none other than the complex term whenever the term “i” is mentioned in mathematics we will take that term as complex number, the complex number “i” has a specific value which is mentioned as $\sqrt{-1}$ in mathematics and form the above given question we need to simplify that term i.e. we need to find a value which is equivalent to ${{\left( \sqrt{-1} \right)}^{29}}$, so to proceed with this type of question we will divide into forms like we can also write the above equation as ${{\left( \sqrt{-1} \right)}^{29}}={{\left( \sqrt{-1} \right)}^{28}}\times \left( \sqrt{-1} \right)$
Now we can also write the above given equation by derivating them even more i.e. ${{\left( \sqrt{-1} \right)}^{29}}={{\left( {{\left( \sqrt{-1} \right)}^{2}} \right)}^{14}}\times \left( \sqrt{-1} \right)$
In this equation we know a power property where when a term has a power and that number also has a power we can multiply those two power to get the final value of that part which is what we used as we know that 14 times 2 is 28 we can put the whole equation as ${{\left( \sqrt{-1} \right)}^{29}}={{\left( {{\left( \sqrt{-1} \right)}^{2}} \right)}^{14}}\times \left( \sqrt{-1} \right)$ which we can also say that ${{\left( i \right)}^{29}}={{\left( {{i}^{2}} \right)}^{14}}\times \left( i \right)$ now we don’t have to multiply I 29 times, we just need to multiply it 2 times and then that value is supposed to be multiplied 14 times and then we will be able to use that as the coefficient of i and then we will multiply it by the remaining “i” that is the single term “i” that has been left.
So now we will calculate ${{i}^{2}}$ which will come out as -1 and then we will substitute this in the main equation from which we will get ${{\left( i \right)}^{29}}={{\left( -1 \right)}^{14}}\times \left( i \right)$ and we can see that the power is positive and how many times 1 is multiplied with itself it will remain the same so we will get the final value as ${{\left( i \right)}^{29}}=i$
So ${{\left( i \right)}^{29}}$ can also be written as $i$.

Note: In the above mentioned question we need to always simplify the whole equation we cannot always solve with powers more than 5, so we need to change it into the most simple equation like we did in this by converting it in the smallest form to easily calculate the whole equation.