Answered

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Perform the indicated operation and write the answer in the form a + bi.
(3 +81) (4-3i)

Sagot :

AǫᴜᴀWɪᴢ

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \: 36 + 23 i[/tex]

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[tex] \large \tt Solution \: : [/tex]

[tex]\qquad \tt \rightarrow \: (3 + 8i)(4 - 3i)[/tex]

[tex]\qquad \tt \rightarrow \: (3 \sdot 4)+ (3 \sdot - 3i) + (8i \sdot4) + (8i \sdot - 3i)[/tex]

[tex]\qquad \tt \rightarrow \: 12 - 9i + 32i - (24 {i}^{2} )[/tex]

[tex]\qquad \tt \rightarrow \: 12 + 23i -( 24 \sdot - 1)[/tex]

[tex]\qquad \tt \rightarrow \: 12 + 23i + 24[/tex]

[tex]\qquad \tt \rightarrow \: 36 + 23i[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

SOLVING

[tex]\Large\maltese\underline{\textsf{A. What is Asked \space}}[/tex]

Perform the indicated operation and write the answer with the form a+bi.

2 numbers given, one of which is complex

[tex]\Large\maltese\underline{\textsf{B. This problem has been solved!\space\space}}[/tex]

Multiply these two numbers, just like you always multiply binomials.

[tex]\bf{(3+8i)(4-3i)}[/tex] | multiply

[tex]\bf{3\times4+3\times(-3i)+8i\times4+8i\times(-3i)}[/tex] | simplify

[tex]\bf{12-9i+32i-24i^2}[/tex] | this can be simplified A LOT

[tex]\bf{12+23i-24i^2}[/tex]  | as strange as it may seem, this can be simplified even  more, because isn't i^2 the same as -1?

[tex]\bf{12+23i-24\times(-1)}=12+23i+24}[/tex] | add 12 and 24

[tex]\bf{36+23i}[/tex]

[tex]\rule{300}{1.7}[/tex]

[tex]\bf{Result:}[/tex]

                     [tex]\bf{=36+23i}[/tex]. The answer is written in the form a+bi, as requested.

[tex]\boxed{\bf{aesthetic\not101}}[/tex]

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