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astra-53 [7]
2 years ago
6

3+5i/-2+3i perform operation

Mathematics
1 answer:
melisa1 [442]2 years ago
4 0

Answer:

\dfrac{9}{13}-\dfrac{19}{13}i

Step-by-step explanation:

Remember i^2=-1

You are given the fraction \dfrac{3+5i}{-2+3i}

First, multiply the numerator and the denominator by -2-3i:

\dfrac{3+5i}{-2+3i}=\dfrac{(3+5i)(-2-3i)}{(-2+3i)(-2-3i)}=\dfrac{(3+5i)(-2-3i)}{(-2)^2-(3i)^2}=\dfrac{(3+5i)(-2-3i)}{4-9i^2}=\dfrac{(3+5i)(-2-3i)}{4+9}

This gives you 13 in denominator, now multiply two complex numbers in numerator:

(3+5i)(-2-3i)=-6-9i-10i-15i^2=-6-19i+15=9-19i

Thus, the initial fraction is

\dfrac{9-19i}{13}=\dfrac{9}{13}-\dfrac{19}{13}i

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