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photoshop1234 [79]
3 years ago
11

Write the quotient as a +bi 4+6i/ 2-3i​

Mathematics
2 answers:
vladimir1956 [14]3 years ago
4 0

\frac{4+6i}{2-3i}=\frac{(4+6i)(2+3i)}{(2-3i)(2+3i)}

\frac{8+12i+12i+18i^2}{4+6i-6i-9i^2}

\frac{-10+24i}{13}

\boxed{\frac{-10}{13}+\frac{24}{13}i}

Kipish [7]3 years ago
4 0

Answer:

- \frac{10}{13} + \frac{24}{13} i

Step-by-step explanation:

Given

\frac{4+6i}{2-3i}

To rationalise the denominator, multiply the numerator/ denominator by the complex conjugate of the denominator.

The conjugate of 2 - 3i is 2 + 3i, thus

\frac{(4+6i)(2+3i)}{(2-3i)(2+3i)} ← expand factors

= \frac{8+24i+18i^2}{4-9i^2} → i² = - 1

= \frac{8+24i-18}{4+9}

= \frac{-10+24i}{13}

= - \frac{10}{13} + \frac{24}{13} i

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