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olganol [36]
3 years ago
13

$i^5+i^{-25}+i^{45}$

^{-25}+i^{45}$" alt="$i^5+i^{-25}+i^{45}$" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
Zina [86]3 years ago
5 0

Answer:

1/i +2i

Step-by-step explanation:

i^5+i^-25+i^45    i^2=-1

i^4*i +i^-24*i^-1 +i^44 *i

(i²)² i+ (i²)^-12*i^-1+(i²)^22 *i     since i²=1

i+i^-1+i=

i+1/i +i=1/i +2i

gregori [183]3 years ago
3 0

i^5+i^{-25}+i^{45}

Rewrite the term i^{-25}

=i^5+\dfrac{1}{i^{25}}+i^{45}

Expand each term so we have

=i(i^2)^2+\dfrac{1}{i(i^2)^{12}}+i(i^2)^{22}

Use the fact that i^2=-1

=i(-1)^2+\dfrac{1}{i(-1)^{12}}+i(-1)^{22}

Use the fact that (-1)^{a}=1 when a is an even number

=i+\dfrac{1}{i}+i

Simplify

=i-i+i

=i

Let me know if you need any clarifications, thanks!

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