Question

What is the value of \[{{\left( -\sqrt{-1} \right)}^{4n+3}}+{{\left( {{i}^{41}}+{{i}^{-257}} \right)}^{9}}\], where \[n\in N\]?

Answer 1

Hint: We know that \[\sqrt{-1}\] is denoted by \[i\]. Thus, \[{{i}^{2}}=-1\]. First we will find a general trend in the higher powers of \[i\] and then will substitute the same in the above equation to be solved. The trend will help us to reduce the higher powers of \[i\] to lower powers of \[i\] and thus help us to solve the problem easily.

Complete step-by-step solution -
We know that, \[i=\sqrt{-1}\].
Squaring on both sides would give us, \[{{i}^{2}}=-1\].
Now, multiplying \[i\] on both sides again would give us \[{{i}^{3}}=-i\].
Consider, \[{{i}^{2}}=-1\] again.
Squaring on both sides would give us \[{{i}^{4}}=1\].
Thus, we got to know that,
\[i=\sqrt{-1}\], \[{{i}^{2}}=-1\], \[{{i}^{3}}=-i\] and \[{{i}^{4}}=1\].
Now let us observe who write \[{{i}^{5}},{{i}^{6}},{{i}^{7}}\] and \[{{i}^{8}}\]. Now, consider \[{{i}^{5}}\]. \[{{i}^{5}}\] can be written as \[{{i}^{4+1}}\]. \[{{i}^{4+1}}\] can be written as \[{{i}^{4}}\times {{i}^{1}}\]. But, \[{{i}^{4}}=1\] as we saw earlier. Thus, \[{{i}^{5}}=i\].
Now, consider \[{{i}^{6}}\]. \[{{i}^{6}}\] can be written as \[{{i}^{4+2}}\].
\[{{i}^{4+2}}\] can be written as \[{{i}^{4}}\times {{i}^{2}}\]. But \[{{i}^{4}}=1\] and \[{{i}^{2}}=-1\].
So, \[{{i}^{6}}=\left( 1 \right)\left( -1 \right)=-1\].
Let us now see \[{{i}^{7}}\]. \[{{i}^{7}}\] can be written as \[{{i}^{4+3}}\]. \[{{i}^{4+3}}\] can be written as \[{{i}^{4}}\times {{i}^{3}}\]. \[{{i}^{4}}=1\] and \[{{i}^{3}}=-i\] as we saw earlier.
Thus, \[{{i}^{7}}=\left( 1 \right)\left( -i \right)=-i\].
Now, \[{{i}^{8}}\] can be written as \[{{i}^{4+4}}\]. \[{{i}^{4+4}}\] can be written as \[{{i}^{4}}\times {{i}^{4}}\]. But, \[{{i}^{4}}=1\]. Thus, \[{{i}^{8}}=\left( 1 \right)\left( 1 \right)=1\].
Therefore to generalize we can write,
\[{{i}^{4n}}=1\], where, \[n\in N\].
Eg: \[{{i}^{4}},{{i}^{8}},{{i}^{12}},{{i}^{14}}\], etc.
\[{{i}^{4n+1}}=i\], where, \[n\in N\].
Eg: \[{{i}^{1}},{{i}^{5}},{{i}^{9}},{{i}^{13}}\], etc.
\[{{i}^{4n+2}}=-1\], where, \[n\in N\].
Eg: \[{{i}^{2}},{{i}^{6}},{{i}^{10}},{{i}^{14}}\], etc.
\[{{i}^{4n+3}}=-i\], where, \[n\in N\].
Eg: \[{{i}^{3}},{{i}^{7}},{{i}^{11}},{{i}^{15}}\], etc.
Now let us take the equation or sum to be solved.
\[{{\left( -\sqrt{-1} \right)}^{4n+3}}+{{\left( {{i}^{41}}+{{i}^{-257}} \right)}^{9}}\]
Now, we know that, \[i=\sqrt{-1}\].
Thus, the sum becomes,
\[{{\left( -i \right)}^{4n+3}}+{{\left( {{i}^{41}}+{{i}^{-257}} \right)}^{9}}\]
Now, this can be written as,
\[={{\left( -1 \right)}^{4n+3}}\times {{\left( i \right)}^{4n+3}}+{{\left( {{i}^{41}}+{{i}^{-257}} \right)}^{9}}\]
Now, \[4n+3\] is always an odd number.
Thus, \[{{\left( -1 \right)}^{4n+3}}=-1\], since odd powers of \[\left( -1 \right)\] give us \[\left( -1 \right)\].
Also, as we saw before, \[{{\left( i \right)}^{4n+3}}=-i\].
Now, let us simplify \[{{i}^{41}}\] and \[{{i}^{-257}}\].
\[{{i}^{41}}\] can be written as \[{{i}^{40+1}}={{i}^{4\left( 10 \right)+1}}\].
Thus, it is of the form, \[{{i}^{4n+1}}\], which is \[i\].
Thus, \[{{i}^{41}}=i\].
Now, \[{{i}^{-257}}\] can be written as \[\dfrac{1}{{{i}^{257}}}\].
Now, \[{{i}^{257}}\] can be written as \[{{i}^{256+1}}\]. \[{{i}^{256+1}}\] can be written as \[{{i}^{4\left( 64 \right)+1}}\] which is of the form \[{{i}^{4n+1}}\], which is \[i\].
Thus, \[{{i}^{257}}=i\].
Also, \[{{i}^{2}}=-1\].
Cross multiplying \[i\] we get, \[i=\dfrac{-1}{i}\].
Multiplying both sides by negative sign we get, \[i=\dfrac{-1}{i}\].
Thus, \[{{i}^{-257}}=\dfrac{1}{{{i}^{257}}}\]
\[\begin{align}
  & {{i}^{-257}}=\dfrac{1}{{{i}^{256+1}}} \\
 & {{i}^{-257}}=\dfrac{1}{i} \\
 & {{i}^{-257}}=\left( -i \right) \\
\end{align}\]
Thus, \[{{\left( -\sqrt{-1} \right)}^{4n+3}}=\left( -1 \right)\left( -i \right)=i\].
\[{{i}^{41}}=i\] and \[{{i}^{-256}}=-i\].
Thus substituting all these in the sum, we get,
\[\begin{align}
  & {{\left( -\sqrt{-1} \right)}^{4n+3}}+{{\left( {{i}^{41}}+{{i}^{-257}} \right)}^{9}}=i+{{\left( i-i \right)}^{9}} \\
 & {{\left( -\sqrt{-1} \right)}^{4n+3}}+{{\left( {{i}^{41}}+{{i}^{-257}} \right)}^{9}}=i+{{\left( 0 \right)}^{9}} \\
 & {{\left( -\sqrt{-1} \right)}^{4n+3}}+{{\left( {{i}^{41}}+{{i}^{-257}} \right)}^{9}}=i+0 \\
 & {{\left( -\sqrt{-1} \right)}^{4n+3}}+{{\left( {{i}^{41}}+{{i}^{-257}} \right)}^{9}}=i \\
\end{align}\]
Thus, \[{{\left( -\sqrt{-1} \right)}^{4n+3}}+{{\left( {{i}^{41}}+{{i}^{-257}} \right)}^{9}}=i\].

Note: You should be very careful with the signs of the complex numbers as in complex numbers squared value comes with a negative which is non – intuitive. This mistake of signs can cause a drastic change in answer.
Example: - If you forgot the negative sign in \[{{i}^{2}}=1\] and write \[{{i}^{2}}=1\], then \[{{i}^{3}}=i\] and \[{{i}^{4}}=1\] which is completely wrong.
Also apply those generalizations of \[{{i}^{4n+1}},{{i}^{4n+2}},{{i}^{4n+3}}\] and \[{{i}^{4n}}\] only if n is a positive number or zero. If you get a negative power, reciprocate it, write it as a fraction, make the power positive and then apply the generalization.