Question

qυєsтıση :-

Evaluate
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$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {\bigg( i^{41} + \dfrac{1}{i^{257}} \bigg) {}^9}$
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Here,
⠀⠀⠀⠀⠀❏ i = iota
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⚠️No spamming⚠️
⚠️Need quality answer⚠️​

Answer 1

The cycle of $i$

$i$ is a number such that $i^2=-1$.

Then, it satisfies $i^4=1$. And then $i^{4k}=1$.

1 is a number that the product equals the previous number.

Hence

• $i^{4k+1}=i$
• $i^{4k+2}=-1$
• $i^{4k+3}=-i$
• $i^{4k}=1$

It cycles every 4th number.

Now, what numbers are $i^{41}$ and $i^{257}$? Let's focus on the remainder.

$41=4\times 10+1$, the remainder of 1, then $i^{41}=i$.

$257=4\times 64+1$, the remainder of 1, then $i^{257}=i$.

Hence,

$(i+\dfrac{1}{i} )^9$

$=(i+\dfrac{1\times i}{i\times i} )^9$

$=(i+\dfrac{i}{-1} )^9$

$=(i-i)^9$

$=0$

And hence, the value of $(i^{47}+\dfrac{1}{i^{257}})^9=0$.

More information

We can find out $\sqrt{i}$ cycles every 8th number using the same logic since $(\sqrt{i} )^8=1$.

Let's calculate the powers.

• $(\sqrt{i} )^{8k+1}=\sqrt{i}$
• $(\sqrt{i} )^{8k+2}=i$
• $(\sqrt{i} )^{8k+3}=i\sqrt{i}$
• $(\sqrt{i} )^{8k+4}=-1$
• $(\sqrt{i} )^{8k+5}=-\sqrt{i}$
• $(\sqrt{i} )^{8k+6}=-i$
• $(\sqrt{i} )^{8k+7}=-i\sqrt{i}$
• $(\sqrt{i} )^{8k}=1$

Adding everything gives 0.

So,

$(\sqrt{i} )+(\sqrt{i} )^{2}+(\sqrt{3} )^{3}+(\sqrt{i} )^{4}+(\sqrt{i} )^{5}+(\sqrt{i} )^{6}+(\sqrt{i} )^{7}+(\sqrt{i} )^{8}=0$

Answer 2

$\large\sf\underline{Given}$

$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {\bigg( i^{41} + \dfrac{1}{i^{257}} \bigg) {}^9}$

$\large\sf\underline{Solution}$

$\sf\:(i^{41}+\frac{1}{i^{257}})^{9}$

Now let's first compute $\sf\:i^{41}$ :

$\sf\:i^{41}$

$\sf⟹\:i^{4 \times 10 + 1}$

$\sf⟹\:(i^{4})^{10} \times i^{1}$

$\sf⟹\:(1)^{10} \times i$

$\sf⟹\:1 \times i$

$\sf⟹\:i$

So $\small{\underline{\boxed{\mathrm\purple{i^{41}=i}}}}$

Now again let's compute $\sf\:i^{257}$ :

$\sf\:\frac{1}{i^{257}}$

$\sf⟹\:\frac{1}{i^{4 \times 64+1}}$

$\sf⟹\:\frac{1}{(i^{4})^{64} \times i^{1}}$

$\sf⟹\:\frac{1}{(1)^{64} \times i^{1}}$

$\sf⟹\:\frac{1}{1 \times i^{1}}$

$\sf⟹\:\frac{1}{i}$

Now dividing and multiplying the expression by i :

$\sf\:\frac{1 \times i}{i \times i}$

$\sf⟹\:\frac{i}{i^{2}}$

$\sf⟹\:\frac{i}{-1}$

$\sf⟹\:-i$

So $\small{\underline{\boxed{\mathrm\purple{i^{257}=-i}}}}$

Now substituting the value of $\sf\:i^{41}$ and $\sf\:i^{257}$ in the given expression we get :

$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {\bigg( i^{41} + \dfrac{1}{i^{257}} \bigg) {}^9}$

$\sf⟹\:(i^{41}+\frac{1}{i^{257}})^{9}$

$\sf⟹\:[i+(-i)]^{9}$

$\sf⟹\:[i-i]^{9}$

$\sf⟹\:[0]^{9}$

$\sf⟹\:0$

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${\sf{{\pink{More\:to\:know:}}}}$

• $\small{\underline{\boxed{\mathrm\red{i=\sqrt{-1}}}}}$

• $\small{\underline{\boxed{\mathrm\red{i^{2}=-1}}}}$

• $\small{\underline{\boxed{\mathrm\red{i^{3}=-i}}}}$

• $\small{\underline{\boxed{\mathrm\red{i^{4}=1}}}}$

‎ ‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎‎ ‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎ ‎ ━━━━━━━━━━━━━━━━

Thnku :)