Question

Show that

{i^19 + (1/i)^25}^2 = -4​

Answer 1

check the attachment. thank you

Answer 2

Answer:

We know that

${i}^{4} = 1$

${i}^{3} = - i$

${i}^{2} = - 1$

${i}^{19} = {i}^{4 \times 4 + 3} = 1 \times {i}^{3} = - i$

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

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

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

-1-1-2

-4

Hence proved