Solution:
[tex]\frac{\sqrt-9}{[(3-2 i)+(1+5 i)]}[/tex]
Remember these complex formulas
1. [tex]\sqrt{-1}=i[/tex]
2. (a + b i) + (c +d i)=(a +c) + i(b+d), i.e real part should be added or subtracted to real part and imaginary part should be added or subtracted to imaginary part.
3. ( a + b i)(a - bi)= a² + b²
4. [tex]i=\sqrt{-1},i^2=-1, i^3= -i, i^4=1[/tex]
→So,[tex]\sqrt{-9}= \sqrt{-1} \times \sqrt{9}= 3 i[/tex]
→3 - 2 i + 1 + 5 i= 4 + 3 i
→[tex]\frac{1}{4+3 i}=\frac{4 - 3 i}{(4 + 3 i)(4 - 3 i)}=\frac {4-3 i}{25}[/tex]→→Rationalizing the Denominator i.e complex number
→[tex]\frac{\sqrt-9}{[(3-2 i)+(1+5 i)]}[/tex]
= [tex]\frac{3 i (4 - 3 i)}{25}=\frac{12 i +9}{25}=\frac{9}{25} +\frac{12 i}{25}[/tex]
