Math, asked by josnapais, 4 months ago

express the complex number i¹⁸+(l/i)²⁵ on the form a+ib​

Answers

Answered by senboni123456
0

Step-by-step explanation:

We have,

z =  {i}^{18} + \frac{1}{i^{25} } \\

 \implies \: z =  {i}^{4 \times 4 + 2}  +  \frac{1}{ {i}^{4 \times 6 + 1} }  \\

 \implies \: z =  - 1 +  \frac{1}{i}  \\

 \implies \: z =  - 1 +  \frac{i}{ {i}^{2} }  \\

 \implies \: z =  - 1 - i

Answered by Anonymous
11

Correct Question :

  • Express the complex number \sf \:  \left(i {}^{18}   + ( \dfrac{1}{i}) {}^{25}  \right) {}^{3} on the form a+ib.

Explanation :

\longrightarrow \sf \:  \left(i {}^{18}   + ( \dfrac{1}{i}) {}^{25}  \right) {}^{3}

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\longrightarrow \sf \bigg( i {}^{28} +   \dfrac{1}{i {}^{25} } \bigg) {}^{3}

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\longrightarrow \sf \:   \bigg((i {}^{2} ) {}^{9}  +  \dfrac{1}{(i)(i {}^{2} ) {}^{12} }  \bigg) {}^{3}

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\because \boxed{  \sf\red{ i {}^{2}  = -  1}}

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\longrightarrow \sf \bigg( ( - 1) {}^{9} +  \dfrac{1}{ i\times ( - 1) {}^{ 12} }  \bigg) {}^{3}

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\longrightarrow \sf \bigg(  - 1 +  \dfrac{1}{i} \bigg) {}^{3}

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\longrightarrow \sf \bigg(  - 1 +  \dfrac{1}{i} \times  \frac{i}{ i}  \bigg) {}^{3}

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\longrightarrow \sf \bigg( - 1 +  \dfrac{i}{i {}^{2} }  \bigg) {}^{3}

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\because \boxed{  \sf\red{ i {}^{2}  = -  1}}

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\longrightarrow \sf \bigg( - 1 +  \dfrac{i}{ - 1}  \bigg) {}^{3}

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\longrightarrow \sf \bigg( - 1 - i \bigg) {}^{3}

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\longrightarrow \sf   - \bigg(1 + i \bigg) {}^{3}

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\because \boxed{ \sf \red{(a + b ) {}^{3} = a {}^{3}  + 3a {}^{2}  b + 3ab {}^{2} + b {}^{3}  }}

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\longrightarrow \sf \:  -  \bigg(1 {}^{3}   + 3 \times 1 {}^{2}  \times i + 3 \times 1 \times i {}^{2} + i {}^{3}  \bigg)

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\longrightarrow \sf \:  -  \bigg(1 + 3i + 3i {}^{2}  + i {}^{2}(i)  \bigg)

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\longrightarrow \sf \:  -  \bigg( 1 + 3i + 3 \times  - 1 +  - 1 \times i\bigg)

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\longrightarrow \sf \:  -  \bigg(1 + 3i - 3 - i \bigg) \\  \\   \\  \longrightarrow \sf \:  -  \bigg(  - 2 + 2i \bigg)

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 \longrightarrow \boxed{ \sf \blue{2 - 2i} }

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