Prove that the value of the determinant -7 5-3i 2 3 -4i 5-3i 8 4-5i 2 3 -4i 4-5i 9 is real-

7 amp; 5+3i amp; ( 2 3 4i ) #160; 5 3i amp; 8 amp; (4+5i) ( 2 3 +4i ) #160; amp; (4 5i) amp; 9 = 7 [72 {4^2 ( ...

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Inequalities Involving Modulus Function

Prove that th...

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Prove that the value of the determinant $∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} - 7 & 5 + 3 i & \frac{2}{3} - 4 i 5 - 3 i & 8 & 4 + 5 i \frac{2}{3} + 4 i & 4 - 5 i & 9 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$ is real.

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⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - 7 & 5 + 3 i & \frac{2}{3} - 4 i 5 - 3 i & 8 & 4 + 5 i \frac{2}{3} + 4 i & 4 - 5 i & 9 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

∣ ∣ ∣ ∣ ∣ \begin{matrix} - 5 & 3 + 5 i & \frac{3}{2} - 4 i 3 - 5 i & 8 & 4 + 5 i \frac{3}{2} + 4 i & 4 - 5 i & 9 \end{matrix} ∣ ∣ ∣ ∣ ∣

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 - 3 i & 3 + 4 i 2 + 3 i & 0 & 4 - 5 i 3 - 4 i & 4 + 5 i & 2 \end{matrix} ⎤ ⎥ ⎦

∣ ∣ ∣ ∣ \begin{matrix} 2 & 7 & 65 3 & 8 & 75 5 & 9 & 86 \end{matrix} ∣ ∣ ∣ ∣ = 0

(1 + i) (1 + 2 i)

\frac{3 + 2 i}{- 2 + i}

\frac{1}{(2 + i)^{2}}

\frac{1 - i}{1 + i}

\frac{(2 + i)^{3}}{2 + 3 i}

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

\frac{2 + 3 i}{4 + 5 i}

\frac{(1 - i)^{3}}{1 - i^{3}}

(1 + 2 i)^{- 3}

\frac{3 - 4 i}{(4 - 2 i) (1 + i)}

(\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{1 - 4 i}{5 + i})

\frac{5 + \sqrt{2} i}{1 - 2 \sqrt{i}}

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