How do you perform the operation and write the result in standard form given ${(2 + 3 i)}^{2} + {(2 - 3 i)}^{2}$ $(2+3i)^2+(2-3i)^2$ ?

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How do you perform the operation and write the result in standard form given ${(2 + 3 i)}^{2} + {(2 - 3 i)}^{2}$ $(2+3i)^2+(2-3i)^2$ ?

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Answer 1 · 2016-11-12T01:58:20

${(2 + 3 i)}^{2} + {(2 - 3 i)}^{2} = - 10$ $(2+3i)^2+(2-3i)^2 = -10$

Explanation:

${(2 + 3 i)}^{2} + {(2 - 3 i)}^{2} = {2^{2} + 2 (2) (3 i) + {(3 i)}^{2}} + {2^{2} - 2 (2) (3 i) + {(3 i)}^{2}}$ $(2+3i)^2+(2-3i)^2 = {2^2+2(2)(3i)+(3i)^2} + {2^2-2(2)(3i)+(3i)^2}$
$:. (2+3i)^2+(2-3i)^2 = {4+12i+9i^2} + {4-12i+9i^2}$
$:. (2+3i)^2+(2-3i)^2 = 8+18i^2$

Now, $i^2 =-1$ , so

$:. (2+3i)^2+(2-3i)^2 = 8-18$
$:. (2+3i)^2+(2-3i)^2 = -10$

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How do you perform the operation and write the result in standard form given ${(2 + 3 i)}^{2} + {(2 - 3 i)}^{2}$ $(2+3i)^2+(2-3i)^2$ ?

1 Answer

Explanation:

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How do you perform the operation and write the result in standard form given (2+3i)^2+(2-3i)^2(2+3i)2+(2−3i)2(2+3i)^2+(2-3i)^2?

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How do you perform the operation and write the result in standard form given ${(2 + 3 i)}^{2} + {(2 - 3 i)}^{2}$ $(2+3i)^2+(2-3i)^2$ ?