Question

The simplest form of (4+5i)^2/(2+3i)^2

Answer 1

Answer:

Step-by-step explanation:

We have,

$\dfrac{\left(4+5\,i\right)^2}{\left(2+3\,i\right)^2}$

$=\dfrac{\left(4\right)^2+\left(5\,i\right)^2+2\cdot4\cdot5\,i}{\left(2\right)^2+\left(3\,i\right)^2+2\cdot2\cdot3\,i}$

$=\dfrac{16+25\left(i\right)^2+40\,i}{4+9\left(i\right)^2+12\,i}$

$=\dfrac{16-25+40\,i}{4-9+12\,i}$

$=\dfrac{-9+40\,i}{-5+12\,i}$

$=\dfrac{\big(-9+40\,i\big)\big(-5-12\,i\big)}{\big(-5+12\,i\big)\big(-5-12\,i\big)}$

$=\dfrac{45-200\,i+108\,i-480\,i^2}{\left(-5\right)^2+\left(12\right)^2}$

$=\dfrac{45-72\,i+480}{25+144}$

$=\dfrac{525-72\,i}{165}$

$=\dfrac{525}{165}-\dfrac{72}{165}\,i$

$=\dfrac{105}{33}-\dfrac{24}{55}\,i$