What is (i/(3-i))(1/(2+3i)) ?

The solution to (i/(3-i))(1/(2+3i)) is 7/130+9/130 i

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simplify (i/(3-i))(1/(2+3i))

simplify

(\frac{i}{3 - i}) (\frac{1}{2 + 3 i})

\frac{7}{130} + \frac{9}{130} i

Solution steps

(\frac{i}{3 - i}) (\frac{1}{2 + 3 i})

Simplify

(\frac{i}{3 - i}) (\frac{1}{2 + 3 i}) : \frac{i}{3 - i} \cdot \frac{1}{2 + 3 i}

= \frac{i}{3 - i} \cdot \frac{1}{2 + 3 i}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{i 1}{( 3 - i ) ( 2 + 3 i )}

Apply rule:

a \cdot 1 = a

i 1 = i

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

(3 - i) (2 + 3 i) = 9 + 7 i

= \frac{i}{9 + 7 i}

Multiply by the conjugate

\frac{9 - 7 i}{9 - 7 i}

= \frac{i ( 9 - 7 i )}{( 9 + 7 i ) ( 9 - 7 i )}

Simplify

\frac{i ( 9 - 7 i )}{( 9 + 7 i ) ( 9 - 7 i )} : \frac{9 i + 7}{130}

= \frac{9 i + 7}{130}

Rewrite in standard complex form:

\frac{7}{130} + \frac{9}{130} i

= \frac{7}{130} + \frac{9}{130} i

x^{2} - 14 x + 35 = 0

4 (x + 6) = x + 3 (x + 5) + 9

5 x^{2} + 15 x

\frac{2 x}{5} = x - 3

- 5 x^{3} + 16 x^{2} - 9

What is (i/(3-i))(1/(2+3i)) ?
The solution to (i/(3-i))(1/(2+3i)) is 7/130+9/130 i