Respuesta :
Let the given complex number
z = x + ix = [tex]\dfrac{5-i}{3+2i}[/tex]
We have to find the standard form of complex number.
Solution:
∴ x + iy = [tex]\dfrac{5-i}{3+2i}[/tex]
Rationalising numerator part of complex number, we get
x + iy = [tex]\dfrac{5-i}{3+2i}\times \dfrac{3-2i}{3-2i}[/tex]
⇒ x + iy = [tex]\dfrac{(5-i)(3-2i)}{3^2-(2i)^2}[/tex]
Using the algebraic identity:
(a + b)(a - b) = [tex]a^{2}[/tex] - [tex]b^{2}[/tex]
⇒ x + iy = [tex]\dfrac{15-10i-3i+2i^2}{9-4i^2}[/tex]
⇒ x + iy = [tex]\dfrac{15-13i+2(-1)}{9-4(-1)}[/tex] [ ∵ [tex]i^{2} =-1[/tex]]
⇒ x + iy = [tex]\dfrac{15-2-13i}{9+4}[/tex]
⇒ x + iy = [tex]\dfrac{13-13i}{13}[/tex]
⇒ x + iy = [tex]\dfrac{13(1-i)}{13}[/tex]
⇒ x + iy = 1 - i
Thus, the given complex number in standard form as "1 - i".
