Question

(sqrt3-sqrt3i)^4

Answer 1

The increasing order of the complex numbers is (√2 - i)⁶ < (√2 - √2i)⁸ = (√3 - i)⁶ =  (-1 + √3i)¹² < (√3 - √3i)⁴.

Absolute values of the complex numbers

The absolute values of the complex numbers are determined as follows;

(sqrt3-sqrt3i)^4 = (√3 - √3i)⁴

$|z| = \sqrt{(\sqrt{3} )^2 + (\sqrt{3 }\times1 )^2} } \\\\|z| = \sqrt{6}$

(-1+sqrt3i)^12 = (-1 + √3i)¹²

$|z| = \sqrt{(-1)^2 + (\sqrt{3)^2} } \\\\|z| = \sqrt{4} \\\\|z| = 2$

(sqrt 3-i)^6 = (√3 - i)⁶

$|z| = \sqrt{(\sqrt{3})^2 + (-1)^2 } \\\\|z| = \sqrt{4} \\\\|z| = 2$

(sqrt2-sqrt2i)^8 = (√2 - √2i)⁸

$|z| = \sqrt{(\sqrt{2} )^2 + (\sqrt{2})^2 } \\\\|z| = 2$

(sqrt2-i)^6 = (√2 - i)⁶

$|z| = \sqrt{(\sqrt{2})^2 + (-1)^2} } \\\\|z| = \sqrt{3}$

Increasing order of the complex numbers;

(√2 - i)⁶ < (√2 - √2i)⁸ = (√3 - i)⁶ =  (-1 + √3i)¹² < (√3 - √3i)⁴.

Learn more about complex numbers here: brainly.com/question/10662770

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