Given:
⇒ a + ib = \(\frac{5+\sqrt2i}{1-\sqrt2i}\)
Multiplying and dividing with 1+√2i
⇒ a + ib = \(\frac{5+\sqrt2i}{1-\sqrt2i} \times \frac{1+\sqrt2i}{1+\sqrt2i}\)
⇒ a + ib = \(\frac{5(1+\sqrt2i)+\sqrt2i(1+\sqrt2i)}{1^2-(\sqrt2i)^2}\)
⇒ a + ib = \(\frac{5+5\sqrt2i+\sqrt2i+2i^2}{1-2i^2}\)
We know that i2=-1
⇒ a + ib = \(\frac{5+6\sqrt2i+2(-1)}{1-2(-1)}\)
⇒ a + ib = \(\frac{3+6\sqrt2i}{3}\)
⇒ a + ib = 1 + 2√2i
∴ The values of a, b are 1, 2√2 .