4 years ago

How do you solve this?

1 answer:

5 0

$\dfrac{3+4i}{2+i}=\dfrac{(3+4i)(2-i)}{(2+i)(2-i)^{(*)}}=\dfrac{3\cdot2-3\cdot i+4i\cdot2-4i\cdot i}{2^2-i^2^{(**)}}\\\\=\dfrac{6-3i+8i+4}{4-(-1)}}=\dfrac{10+5i}{4+1}=\dfrac{10+5i}{5}=\dfrac{10}{5}+\dfrac{5i}{5}=\boxed{2+i}\to \boxed{a=2}\\\\used:\\(*)\ (a+b)(a-b)=a^2-b^2\\(**)\ i^2=-1$

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My name is Ann [436]

Answer:

Step-by-step explanation:

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So,

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3 years ago

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WARRIOR [948]

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3 years ago

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3 years ago

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krek1111 [17]

Answer:

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Step-by-step explanation:

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divide by 5

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