Answer:
The complex numbers arranged in increasing order of their absolute value (modulus) are:
Step-by-step explanation:
-5(n³-n²-1)+ n(n²-n)
= -5n³+5n²+5+ n³-n² (multiplying)
= -6n³+6n²+5
(n²-1)(n +2) -n² (n-3)
= n³+2n²-n -2 - n³+ 3n² (multiplying)
=5n²-n -2
n² (n-4) + 5n³ -6
= n³- 4n +5n³ -6 (multiplying)
= 6n³- 4n -6
2n( n²-2n-1) + 3n²
= 2n³- 4n² -2n + 3n²
=2n³ +n² -2n
First we solve the expression by multiplying adding or subtracting . Then we take the expression whose answer is the highest both in co efficient and the power to be the last in number as we have to write the expression in increasing order , mso we will start with the smallest.
The expression whose answer is the smallest both in co efficient and the power is taken to be thefirst in number as we have to write the expression in increasing order. The negative co efficients and less powers are taken to be the smallest in value.
The complex numbers in increasing order of their absolute value is [tex](\sqrt 3 - \sqrt 3i)^4[/tex] and [tex](\sqrt 3 - i)^6[/tex]
The complex numbers are given as:
[tex](\sqrt 3 - \sqrt 3i)^4[/tex] and [tex](\sqrt 3 - i)^6[/tex]
Start by simplifying both complex numbers using binomial expansion method.
So, we have:
[tex](\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[/tex]
Evaluate the exponents
[tex](\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[/tex]
Evaluate the products
[tex](\sqrt 3 - \sqrt 3i)^4 = 9 + 12\sqrt 3 \times \sqrt 3i - 54 - 12i \times (\sqrt 3) \times \sqrt 3 + 9[/tex]
Evaluate the products
[tex](\sqrt 3 - \sqrt 3i)^4 = 9 + 36i - 54 - 36i + 9[/tex]
Evaluate like terms
[tex](\sqrt 3 - \sqrt 3i)^4 = 9 - 54 + 9[/tex]
[tex](\sqrt 3 - \sqrt 3i)^4 = -36[/tex]
Next, simplify [tex](\sqrt 3 - i)^6[/tex]
[tex](\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[/tex]
Simplify
[tex](\sqrt 3 - i)^6 = -64[/tex]
So, we have:
[tex](\sqrt 3 - \sqrt 3i)^4 = -36[/tex] and [tex](\sqrt 3 - i)^6 = -64[/tex]
Calculate the absolute values
[tex]|(\sqrt 3 - \sqrt 3i)^4 |= |-36|[/tex]
[tex]|(\sqrt 3 - \sqrt 3i)^4 |= 36[/tex]
[tex]|(\sqrt 3 - i)^6 |= |-64|[/tex]
[tex]|(\sqrt 3 - i)^6 |= 64[/tex]
By comparison: 36 is less than 64
So, the complex numbers in increasing order of their absolute value is [tex](\sqrt 3 - \sqrt 3i)^4[/tex] and [tex](\sqrt 3 - i)^6[/tex]
Read more about complex numbers at:
https://brainly.com/question/10662770
