Given:
⇒ a +ib = \(\frac{2+3i}{4+5i}\)
Multiplying and dividing with 4-5i
⇒ a +ib = \(\frac{2+3i}{4+5i}\) x \(\frac{4-5i}{4-5i}\)
⇒ a +ib = \(\frac{2(4-5i)+3i(4-5i)}{(4)^2-(5i)^2}\)
⇒ a +ib = \(\frac{8-10i+12i-15i^2}{16-25i^2}\)
We know that i2=-1
⇒ a +ib = \(\frac{8+2i-15(-1)}{16-25(-1)}\)
⇒ a +ib = \(\frac{23+2i}{41}\)
∴ The values of a, b are \(\frac{23}{41}\) , \(\frac{2}{41}\) .