Question

50 points

prove that:-

${( {i}^{19} + ( \frac{1}{ {i}^{25} } ))}^{2} = - 4$
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Answer 1

we know that

{i}^{2} = - 1 \\ {i}^{4} =1 \\

so multiple of 4 iota's value is 1.

{i}^{19} = {i}^{4 + 4 + 4 + 4 + 3} = {i}^{3} = - i

{( \frac{1}{i} })^{25} = { \frac{1}{ {i}^{6 \times 4 + 1} } }

= \frac{1}{i} \\

( { - i + \frac{1}{i} })^{2} \\ = {( \frac{ - {i}^{2} + 1}{i} })^{2} = ( { \frac{2}{i} })^{2}

= \frac{4}{ {i}^{2} } \\ = - 4 \:

PLZ REFER THE IMAGE.....

Answer 2

Prerequisite knowledge:

Before jumping on to the question your should know some basic formulas and results

1.(a+b)²=a²+2ab+b²

(The above formula can be proved by binomial expansion of (a+b)ⁿ)

2.i=√(-1)

3.i²=-1

4.i³=-i

5.i⁴=1

To prove:

${( {i}^{19} + ( \frac{1}{ {i}^{25} } ))}^{2} = - 4$

Proof:

i¹⁹=(i⁴)⁴*i³

=1⁴*(-i)

=-i

i²⁵=(i⁴)⁶*i

=1⁶*i

=i

From the above simplifications we can re write the equation as

${( {-i} + ( \frac{1}{ {i} } ))}^{2}$

Using (a+b)² formula we get

${( {-i} + ( \frac{1}{ {i} } ))}^{2}\\=(-i)^2 + (\frac{1}{i})^2 + 2*(-i)*(\frac{1}{i})\\=i^2 + \frac{1}{i^2} + 2(-1)\\=-1 -1 -2\\=-4$

Hence proved

Hope this helps.