Prove that \[{{i}^{-35}}=i\].

Hint: This question belongs to the topic of complex numbers. In this question, we will first understand the value and the symbol of the term iota. After that, we will find out the values of \[i\], \[{{i}^{2}}\], \[{{i}^{3}}\], and \[{{i}^{4}}\]. After that, we will find the value of the inverse of iota. After that, we will prove that \[{{i}^{-35}}=i\] on solving the further process.

Complete step by step answer:
Let's solve this question.
In this question, we have asked to prove that \[{{i}^{-35}}=i\].
Let us first know what iota is.
The symbol for the term iota is \[i\] and the value of the term iota is square root of negative one.
So, we can write
\[i=\sqrt{-1}\]
So, the square of iota will be
\[{{i}^{2}}=i\times i={{\left( \sqrt{-1} \right)}^{2}}=-1\]
The cube of iota will be
\[{{i}^{3}}={{i}^{2}}\times i=\left( -1 \right)\times i=-i\]
The fourth power of iota will be
\[{{i}^{4}}={{i}^{2}}\times {{i}^{2}}=\left( -1 \right)\times \left( -1 \right)=1\]
Now, let us find out the value of inverse of iota that is \[\dfrac{1}{i}\]
Hence, we can write the inverse of iota as
\[{{i}^{-1}}=\dfrac{1}{i}\]
After multiplying iota on numerator and denominator in the right side of the above equation, we get
\[\Rightarrow {{i}^{-1}}=\dfrac{1}{i}\times \dfrac{i}{i}=\dfrac{i}{i\times i}=\dfrac{i}{{{i}^{2}}}\]
As we know that the value of the square of iota is -1, so we can write
\[\Rightarrow {{i}^{-1}}=\dfrac{i}{-1}=-i\]
Hence, we get that the inverse of iota is negative of iota.
Now, let us solve for \[{{i}^{-35}}\].
As we know that, \[{{\left( x \right)}^{a+b}}\] can also be written as \[{{\left( x \right)}^{a+b}}={{x}^{a}}\times {{x}^{b}}\], so we can write the term \[{{i}^{35}}\] as
\[{{i}^{35}}={{i}^{\left( 4+4+4+4+4+4+4+4+4-1 \right)}}={{i}^{4}}\times {{i}^{4}}\times {{i}^{4}}\times {{i}^{4}}\times {{i}^{4}}\times {{i}^{4}}\times {{i}^{4}}\times {{i}^{4}}\times {{i}^{4}}\times {{i}^{-1}}\]
Putting the value of \[{{i}^{4}}\] as 1 in the above equation, we get
\[\Rightarrow {{i}^{35}}=1\times 1\times 1\times 1\times 1\times 1\times 1\times 1\times 1\times {{i}^{-1}}={{i}^{-1}}\]
\[\Rightarrow {{i}^{35}}={{i}^{-1}}\]
Now, by multiplying -1 in the power to the both side of equation, we get
\[\Rightarrow {{\left( {{i}^{35}} \right)}^{-1}}={{\left( {{i}^{-1}} \right)}^{-1}}\]
Using the formula \[{{\left( {{x}^{a}} \right)}^{b}}={{x}^{a\times b}}\], we can write in the above equation as
\[\Rightarrow {{i}^{-35}}=i\]
Hence, we have proved that \[{{i}^{-35}}=i\].

Note: We can solve this question using an alternate method. Whenever we have to find out the value of iota having greater powers, then we will make that term of iota having greater powers in the form of \[{{i}^{4n+r}}\]. For that, we should remember that \[i=\sqrt{-1}\], \[{{i}^{2}}=-1\], \[{{i}^{3}}=-i\], and \[{{i}^{4}}=1\].
So, we can write \[{{i}^{35}}\] as
\[{{i}^{\left( 35 \right)}}={{i}^{\left( 4\times 8+3 \right)}}={{i}^{\left( 4\times 8 \right)}}{{i}^{\left( 3 \right)}}={{\left( {{i}^{\left( 4 \right)}} \right)}^{8}}{{i}^{\left( 3 \right)}}\]
Using the formula \[{{i}^{3}}=-i\] and \[{{i}^{4}}=1\], we can write
\[\Rightarrow {{i}^{\left( 35 \right)}}={{\left( 1 \right)}^{8}}\left( -i \right)=-i\]
Multiplying -1 to the power in both the sides of the equation, we get
\[\Rightarrow {{i}^{\left( -35 \right)}}={{\left( -i \right)}^{-1}}=-{{\left( i \right)}^{-1}}\]
As we know that inverse of iota is negative of iota, so we can write
\[\Rightarrow {{i}^{-35}}=i\]