What is the answer to z= 1/((1+i)(1-2i)(1+3i)) ?

The answer to z= 1/((1+i)(1-2i)(1+3i)) is z= 3/50-i 2/25

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z= 1/((1+i)(1-2i)(1+3i))

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

z = \frac{3}{50} - i \frac{2}{25}

Solution steps

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

Simplify

\frac{1}{( 1 + i ) ( 1 - 2 i ) ( 1 + 3 i )} : \frac{3}{50} - i \frac{2}{25}

z = \frac{3}{50} - i \frac{2}{25}

3 x^{2} + 4 x y - 4 y^{2}

\frac{1}{4} < \frac{x}{5} < \frac{2}{3}

6 a + 3 = 45

\frac{2 ^{n + 1}}{2 ^{n}}

x + 1 \cdot x - 1

What is the answer to z= 1/((1+i)(1-2i)(1+3i)) ?
The answer to z= 1/((1+i)(1-2i)(1+3i)) is z= 3/50-i 2/25