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erastova [34]
3 years ago
10

(4 + 4i)/(5+4i) = divide

Mathematics
1 answer:
LekaFEV [45]3 years ago
7 0

Answer:

B.

Step-by-step explanation:

\frac{4 + 4i}{5 + 4i}

  • Multiplying both numerator and denominator by (5 - 4i) <em>,</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>c</em><em>o</em><em>n</em><em>j</em><em>u</em><em>g</em><em>a</em><em>t</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>d</em><em>e</em><em>n</em><em>o</em><em>m</em><em>i</em><em>n</em><em>a</em><em>t</em><em>o</em><em>r</em><em>,</em><em> </em><em>i</em><em>.</em><em> </em><em>e</em><em>,</em><em> </em><em>(</em><em>5</em><em> </em><em>+</em><em> </em><em>4</em><em>i</em><em>)</em><em>.</em>

\frac{4 + 4i}{5 + 4i}  \times  \frac{5 - 4i}{5-4i}

\frac{(4 + 4i)(5 - 4i)}{(5 + 4i)(5 - 4i)}

  • Multiplying (4+4i) and (5-4i) using distributive property
  • Using the identity <em>(a+b)(a-b)= a² - b²</em> where 5 will act as a and 4i will act as b

\frac{20-16i+20i-16i^2}{(5) {}^{2} -  (4i) {}^{2} }

  • i² = -1

(<em>combining like terms)</em>

\frac{20+(-16i+20i)-(-16)}{25-(-16)}

\frac{(20+16)+4i}{25+16}

\frac{36+4i}{41}

distributing the denominator

\frac{36}{41} + \frac{4}{41}i

That is, option B.

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