Question

(2+3i)/(4-5i)

Answer 1

(2 + 3i) / (4-5i)

start by rationalizing the denominator (i.e. multiply the top and bottom by (-5i +4)

Answer 2

That's what I thought, but doesn't that leave an I on the bottom?Let me try it.

Answer 3

Multiply the fraction by

\(\dfrac{4 + 5i}{4 + 5i} \)

Answer 4

You need to multuiply the numerator and denominato by the complex conjugate of the denominator.
The complex conjugate of a + bi is a - bi.
The complex conjugate fo 4 - 5i is 4 + 5i.
That's what @blurbendy meant.

Answer 5

mathhstudent55 is correct, it should be 5i + 4, not -5i

Answer 6

Ohh, multiply by the opposite, so by 5i-4

Answer 7

What is i time i again?

Answer 8

-1

Answer 9

\( i = \sqrt{-1} \)

Answer 10

No. You muliply the numerator and denominator by 4 + 5i.

Answer 11

\( \dfrac{2 + 3i}{4 - 5i} \times \dfrac{4 + 5i}{4 + 5i} \)

Answer 12

It is 5i+4 , thanks math student. The answer is 22i-7

Answer 13

\( \dfrac{2 + 3i}{4 - 5i} \times \dfrac{4 + 5i}{4 + 5i} \)

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

\(= \dfrac{8 + 10i + 12i + 15i^2}{16 + 20i - 20i - 25i^2} \)

\( = \dfrac{8 + 22i - 15}{16 + 25} \)

\(= \dfrac{-7 + 22i}{41} \)

Answer 14

\( = -\dfrac{7}{41} + \dfrac{22}{41}i \)

You did the numerator correctly, but you also need to multiply out the denominator.
The denominator doesn't just disappear. The denominator just becomes a real number.