Let the given complex number
z = x + ix =
![(5-i)/(3+2i)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/pxlzxpmmreqgmijei3v81tqxxwlvu944f4.png)
We have to find the standard form of complex number.
Solution:
∴ x + iy =
![(5-i)/(3+2i)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/pxlzxpmmreqgmijei3v81tqxxwlvu944f4.png)
Rationalising numerator part of complex number, we get
x + iy =
![(5-i)/(3+2i)* (3-2i)/(3-2i)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/1t3qru45kx8yb03nk9t3m0j30c7b7ayflk.png)
⇒ x + iy =
![((5-i)(3-2i))/(3^2-(2i)^2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/5mc89js1agdl0dacxtpgxu5bw1pq3gfl5o.png)
Using the algebraic identity:
(a + b)(a - b) =
-
![b^(2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/oau7saf5rpil53q02wg0vrlk1rczrycs5h.png)
⇒ x + iy =
![(15-10i-3i+2i^2)/(9-4i^2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/lflvsa34g15ov7fvy1fsmbmszfaw4ehzjf.png)
⇒ x + iy =
[ ∵
]
⇒ x + iy =
⇒ x + iy =
⇒ x + iy =
⇒ x + iy = 1 - i
Thus, the given complex number in standard form as "1 - i".