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VashaNatasha [74]
2 years ago
10

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

Mathematics
2 answers:
kkurt [141]2 years ago
8 0

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: 36 + 23 i

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \: (3 + 8i)(4 - 3i)

\qquad \tt \rightarrow \: (3 \sdot 4)+ (3 \sdot - 3i) + (8i \sdot4) + (8i \sdot - 3i)

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

\qquad \tt \rightarrow \: 12 + 23i -( 24 \sdot - 1)

\qquad \tt \rightarrow \: 12 + 23i + 24

\qquad \tt \rightarrow \: 36 + 23i

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Savatey [412]2 years ago
8 0
<h2>SOLVING</h2>

\Large\maltese\underline{\textsf{A. What is Asked \space}}

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

2 numbers given, one of which is complex

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

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

\bf{(3+8i)(4-3i)} | multiply

\bf{3\times4+3\times(-3i)+8i\times4+8i\times(-3i)} | simplify

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

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

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

\bf{36+23i}

\rule{300}{1.7}

\bf{Result:}

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

\boxed{\bf{aesthetic\not101}}

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