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, ${{i}^{-257}}=\dfrac{1}{{{i}^{257}}}$

& {{i}^{-257}}=\dfrac{1}{{{i}^{256+1}}} \\\ & {{i}^{-257}}=\dfrac{1}{i} \\\ & {{i}^{-257}}=\left( -i \right) \\\ \end{aligned}$$ Thus, $${{\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{aligned} & {{\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{aligned}$$ Thus, $${{\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.