Complex Arithmetic/Examples/(2i-1)^2 (4(1-i)^-1 + (2-i) (1+i)^1)

Example of Complex Arithmetic

$\paren {2 i - 1}^2 \paren {\dfrac 4 {1 - i} + \dfrac {2 - i} {1 + i} } = -\dfrac {11} 2 - \dfrac {23} 2 i$

Proof

\(\ds \paren {2 i - 1}^2 \paren {\dfrac 4 {1 - i} + \dfrac {2 - i} {1 + i} }\)

\(=\)

\(\ds \paren {\paren {-1}^2 - 2 \times 2 i + 4 i^2}^2 \paren {\dfrac {4 \paren {1 + i} } {\paren {1 - i} \paren {1 + i} } + \dfrac {\paren {2 - i} \paren {1 - i} } {\paren {1 + i} \paren {1 - i} } }\)

\(\ds \)

\(=\)

\(\ds \paren {-3 - 4 i} \paren {\dfrac {4 + 4 i} {1^2 + 1^2} + \dfrac {\paren {2 - i} \paren {1 - i} } {1^2 + 1^2} }\)

\(\ds \)

\(=\)

\(\ds \paren {-3 - 4 i} \paren {\dfrac {4 + 4 i} 2 + \dfrac {2 - 2 i - i + i^2} 2}\)

\(\ds \)

\(=\)

\(\ds \paren {-3 - 4 i} \paren {\dfrac {4 + 4 i} 2 + \dfrac {1 - 3 i} 2}\)

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {\paren {-3} \times 5 - \paren {-4} \times 1} + \paren {\paren {-3} \times 1 + \paren {-4} \times 5} i} 2\)

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {-15 + 4} + \paren {-3 - 20} i} 2\)

\(\ds \)

\(=\)

\(\ds -\dfrac {11} 2 - \dfrac {23} 2 i\)

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $53 \ \text {(h)}$