Question

1 + iota ki power 4 upon 1 minus iota cube whole power 4​

Answer 1

Answer ⇒ -4

Explanation ⇒ Given Question,

   $(\frac{1 + i^4}{1 - i^3})^{4}$

We know, i = √(-1)

∴ i² = -1 and thus, i⁴ = 1

∴ i³ = i².i = -i

Putting these values in question.

We will get,  $(\frac{1 + i^4}{1 - i^3})^{4}$ =

$(\frac{1 + i^4}{1 - i^3})^{4}$ = $(\frac{1 + 1}{1 - (-i)})^{4}\\(\frac{2}{1 + i})^{4}\\(\frac{4}{1 + i^2 + 2i})^2\\(\frac{4}{1 -1 + 2i})^2$

= $(\frac{16}{4i^2})$

= $(\frac{4}{i^2})$

= -4

Hence, the value of the given expression is -4.

Hope it helps.

Answer 2

Answer:

$(\frac{(1 + i)^4}{(1-i)^3})^4$ = -4

Step-by-step explanation:

$(\frac{(1 + i)^4}{(1-i)^3})^4$

Numerator

= (1 + i)⁴

= (( 1 + i)²)²

= ( 1 + i² + 2i)²

= (1 - 1 + 2i)²

= (2i)²

= 4 i²

= 4 (-1)

= -4

Denominator

(1 - i)³

= (1 -i)³ (1 - i)/(1-i)

= (1 -i)⁴/(1-i)

=  ((1 -i)²)²/(1-i)

= (1 + i² - 2i)²/(1 - i)

= 4i²/(1-i)

= -4/(1 - i)

Numerator /Denominator =   (1 - i)

(1 -i)⁴

= ((1 -i)²)²

= (i² + 1 - 2i)²

= 4i²

= -4

$(\frac{1 + i^4}{1 - i^3})^4\\i^4 = 1 , i^3 = -i\\(\frac{2}{1 + i} )^4\\(\frac{2(1 -i)}{(1 + i)(1-i)} )^4\\(\frac{2(1 -i)}{(1 - i^2)} )^4\\(\frac{2(1 -i)}{2} )^4\\(1 - i)^4$

= (( 1 - i)²)²

= ( 1 + i² - 2i)²

= (-2i)²

= 4i²

= -4