Complex Arithmetic/Examples/Imaginary (z 1 z 2 over z 3)

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Example of Complex Arithmetic

Let $z_1 = 1 - i$, $z_2 = -2 + 4 i$ and $z_3 = \sqrt 3 - 2 i$.

Then:

$\map \Im {\dfrac {z_1 z_2} {z_3} } = \dfrac {6 \sqrt 3 + 4} 7$


Proof

\(\ds \map \Im {\dfrac {z_1 z_2} {z_3} }\) \(=\) \(\ds \map \Im {\dfrac {\paren {1 - i} \paren {-2 + 4 i} } {\sqrt 3 - 2 i} }\)
\(\ds \) \(=\) \(\ds \map \Im {\dfrac {\paren {-2 + 2 i + 4 i - 4 i^2} } {\sqrt 3 - 2 i} }\)
\(\ds \) \(=\) \(\ds \map \Im {\dfrac {2 + 6 i} {\sqrt 3 - 2 i} }\)
\(\ds \) \(=\) \(\ds \map \Im {\dfrac {\paren {2 + 6 i} \paren {\sqrt 3 + 2 i} } {\paren {\sqrt 3 - 2 i} \paren {\sqrt 3 + 2 i} } }\)
\(\ds \) \(=\) \(\ds \map \Im {\dfrac {2 \sqrt 3 + 4 i + 6 \sqrt 3 i + 12 i^2} {3 + 2^2} }\)
\(\ds \) \(=\) \(\ds \map \Im {\dfrac {2 \sqrt 3 - 12 + \paren {6 \sqrt 3 + 4} i} 7}\)
\(\ds \) \(=\) \(\ds \dfrac {6 \sqrt 3 + 4} 7\)

$\blacksquare$


Sources