What is ((2+i)/(3+4i)-(2i)/(3-4i))^2 ?

The solution to ((2+i)/(3+4i)-(2i)/(3-4i))^2 is 203/625-396/625 i

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simplify ((2+i)/(3+4i)-(2i)/(3-4i))^2

simplify

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

\frac{203}{625} - \frac{396}{625} i

Solution steps

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

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

(\frac{2 + i}{3 + 4 i} - \frac{2 i}{3 - 4 i})^{2} = (\frac{2 + i}{3 + 4 i})^{2} - 2 \cdot \frac{2 + i}{3 + 4 i} \cdot \frac{2 i}{3 - 4 i} + (\frac{2 i}{3 - 4 i})^{2}

= (\frac{2 + i}{3 + 4 i})^{2} - 2 \cdot \frac{2 + i}{3 + 4 i} \cdot \frac{2 i}{3 - 4 i} + (\frac{2 i}{3 - 4 i})^{2}

Simplify

(\frac{2 + i}{3 + 4 i})^{2} - 2 \cdot \frac{2 + i}{3 + 4 i} \cdot \frac{2 i}{3 - 4 i} + (\frac{2 i}{3 - 4 i})^{2} : \frac{203}{625} - \frac{396}{625} i

= \frac{203}{625} - \frac{396}{625} i

\frac{1}{3} (x + 3) + \frac{1}{2} \geq - \frac{9}{2}

2 (1 - x) \geq 5 x

a^{2} - 4 a - 21

\frac{1}{( x + h )} - \frac{1}{x}

z^{2} + 9 z + 14

What is ((2+i)/(3+4i)-(2i)/(3-4i))^2 ?
The solution to ((2+i)/(3+4i)-(2i)/(3-4i))^2 is 203/625-396/625 i