How do you write $(3 + 2 i) + (5 - i) + 6 i$ $(3+2i)+(5-i)+6i$ in standard form?

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How do you write $(3 + 2 i) + (5 - i) + 6 i$ $(3+2i)+(5-i)+6i$ in standard form?

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Answer 1 · 2016-11-08T15:34:10

$7 i + 8$ $7i + 8$

Explanation:

There is no need of brackets while adding or subtracting
$3 + 2 i + 5 - i + 6 i$ $3 + 2i + 5 - i + 6i$
$2 i - i + 6 i + 8$ $2i - i + 6i + 8$
$i (2 - 1 + 6) + 8$ $i( 2 - 1 + 6) + 8$
$7 i + 8$ $7i + 8$

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How do you write $(3 + 2 i) + (5 - i) + 6 i$ $(3+2i)+(5-i)+6i$ in standard form?

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How do you write (3+2i)+(5-i)+6i(3+2i)+(5−i)+6i(3+2i)+(5-i)+6i in standard form?

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How do you write $(3 + 2 i) + (5 - i) + 6 i$ $(3+2i)+(5-i)+6i$ in standard form?