Question

If $z=\frac{5+\sqrt{2}i }{1-\sqrt{2}i}$, then Re Z is

Answer 1

Answer:

Step-by-step explanation:

We have,

$z=\dfrac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}$

$\implies\,z=\dfrac{\left(5+\sqrt{2}\,i\right)\left(1+\sqrt{2}\,i\right)}{\left(1-\sqrt{2}\,i\right)\left(1+\sqrt{2}\,i\right)}$

$\implies\,z=\dfrac{\left(5+\sqrt{2}\,i+5\sqrt{2}\,i+2\,i^2\right)}{\left(1\right)^2-\left(\sqrt{2}\,i\right)^2}$

$\implies\,z=\dfrac{5+6\sqrt{2}\,i-2}{1+2}$

$\implies\,z=\dfrac{3+6\sqrt{2}\,i}{3}$

$\implies\,z=1+2\sqrt{2}\,i$

$\therefore\,\tt{Re(z)=1}$