Respuesta :

dalendrk
[tex]\dfrac{3-i}{3+i}=\dfrac{(3-i)(3-i)}{(3+i)(3-i)^{(*)}}=\dfrac{(3)(3)-(3)(i)-(i)(3)+(i)(i)^{(**)}}{3^2-i^2^{(**)}}\\\\=\dfrac{9-3i-3i-1}{9-(-1)}=\dfrac{8-6i}{9+1}=\dfrac{8-6i}{10}=\dfrac{2(4-3i)}{2\cdot5}=\boxed{\dfrac{4-3i}{5}}\\\\\boxed{\boxed{\frac{4}{5}-\frac{3}{5}i}}[/tex]

Answer with explanation:

[tex]\frac{3-i}{3+i}\\\\=\frac{(3-i)\times(3-i)}{(3+i)\times(3-i)}\\\\=\frac{(3-i)^2}{3^2-(i)^2}\\\\=\frac{3^2+i^2-2 \times 3 \times i}{9-(-1)}\\\\=\frac{9-1-6 i}{9+1}\\\\=\frac{8-6 i}{10}\\\\=\frac{8}{10}-\frac{6 i}{10}\\\\=\frac{4}{5}-\frac{3i}{5}[/tex]

⇒To write a complex number in standard form,having complex number in it's denominator multiply by it's conjugate in numerator and denominator

Used the following identities to solve the complex number in fractional form

→  (a+bi)(a-bi)=a²+b²

→i=√-1,   i²= -1

[tex]\rightarrow{\overline{a+bi}}=a-b i\\\\\rightarrow{\overline{a-bi}}=a+bi[/tex]

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