Question

please solve this question.
SIMPLIFY IT PLEASE
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Answer 1

Answer : 54√5 - 406i - 45

Step by step explanation :

$( \sqrt{5} + 7i) \times ( \sqrt{5} - 7i) {}^{2} + ( - 2 + 7i) {}^{2}$

Write the exponential as a multiplication.

$= > ( \sqrt{5} + 7i) \times ( \sqrt{5} - 7i) \times ( \sqrt{5} - 7i) + ( - 2 + 7i) {}^{2}$

Use the commutative property to reorder the terms.

$= > ( \sqrt{5} + 7i) \times ( \sqrt{5} - 7i) \times ( \sqrt{5} - 7i) + (7i - 2) {}^{2}$

Using

$(a + b)(a - b) = a {}^{2} - b {}^{2}$

, simplify the product.

$= > (5 - 49i {}^{2} ) \times ( \sqrt{5 } - 7i) + (7i - 2) {}^{2}$

Using

$(a - b) {}^{2} = a { }^{2} - 2ab + b {}^{2}$

expand the expression

$= > (5 - 49i) {}^{2} \times ( \sqrt{5} - 7i) + 49i {}^{2} - 28i + 4$

By definition, i^2 = - 1

$= > (5 - 49 \times ( - 1)) \times ( \sqrt{5} - 7i) + 49 \times ( - 1) - 28i + 4$

Any expression multiplied by - 1 equals its opposite

$= > (5 + 49) \times ( \sqrt{5} - 7i) - 49 - 28i + 4$

Add the numbers

$= > 54( \sqrt{5} - 7i) - 49 - 28i + 4$

Distribute 54 through the parenthesis

$= > 54 \sqrt{5} - 378i - 49 - 28i + 4$

Collect the like terms by calculating the sum or difference of their coefficients

$= > 54 \sqrt{5} - 406i - 49 + 4$

Calculate the sum. Keep the sign of the number with the larger absolute value and subtract the smaller absolute value from the larger

$= > 54 \sqrt{5} - 406i - (49 - 4)$

Subtract the numbers

$= > 54 \sqrt{5} - 406i - 45$