Mathematics, 04.12.2020 17:00 kenzie19169

Simplify the complex numbers and arrange them in increasing order of their absolute value (modulus).

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Answer from: carthur71

The complex numbers in increasing order of their absolute value is $(\sqrt 3 - \sqrt 3i)^4$ and $(\sqrt 3 - i)^6$

The complex numbers are given as:

$(\sqrt 3 - \sqrt 3i)^4$ and $(\sqrt 3 - i)^6$

Start by simplifying both complex numbers using binomial expansion method.

So, we have:

$(\sqrt 3 - \sqrt 3i)^4 = 1 \times (\sqrt 3)^4 + 4 \times (\sqrt 3)^3 \times (\sqrt 3i) + 6 \times (\sqrt 3)^2 \times (\sqrt 3i)^2 + 4 \times (\sqrt 3)^1 \times (\sqrt 3i)^3 + 1 \times (\sqrt 3i)^4$

Evaluate the exponents

$(\sqrt 3 - \sqrt 3i)^4 = 1 \times 9 + 4 \times 3\sqrt 3 \times \sqrt 3i - 6 \times 3 \times 3 \times 1 - 4 \times (\sqrt 3) \times 3i\sqrt 3 + 1 \times 9$

Evaluate the products

$(\sqrt 3 - \sqrt 3i)^4 = 9 + 12\sqrt 3 \times \sqrt 3i - 54 - 12i \times (\sqrt 3) \times \sqrt 3 + 9$

Evaluate the products

$(\sqrt 3 - \sqrt 3i)^4 = 9 + 36i - 54 - 36i + 9$

Evaluate like terms

$(\sqrt 3 - \sqrt 3i)^4 = 9 - 54 + 9$

$(\sqrt 3 - \sqrt 3i)^4 = -36$

Next, simplify $(\sqrt 3 - i)^6$

$(\sqrt 3 - i)^6 = 1 \times (\sqrt 3)^6 + 6 \times (\sqrt 3)^5 \times (i) +15\times (\sqrt 3)^4 \times (i)^2 + 20\times (\sqrt 3)^3 \times (i)^3 + 15 \times (\sqrt 3)^2 \times (i)^4+ 6 \times (\sqrt 3) \times (i)^5 + 1 \times (i)^6$

Simplify

$(\sqrt 3 - i)^6 = -64$

So, we have:

$(\sqrt 3 - \sqrt 3i)^4 = -36$ and $(\sqrt 3 - i)^6 = -64$

Calculate the absolute values

$|(\sqrt 3 - \sqrt 3i)^4 |= |-36|$

$|(\sqrt 3 - \sqrt 3i)^4 |= 36$

$|(\sqrt 3 - i)^6 |= |-64|$

$|(\sqrt 3 - i)^6 |= 64$

By comparison: 36 is less than 64

So, the complex numbers in increasing order of their absolute value is $(\sqrt 3 - \sqrt 3i)^4$ and $(\sqrt 3 - i)^6$

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