Question

Simplify the expression The root of negative nine over the quantity of three minus two i plus the quantity of one plus five i.

Answer 1

This will be:
[√9]/[(3-2i)+(1+5i)]
simplifying  the above we get:
√9/(3+1-2i+5i)
=√9/(4+3i)
rationalizing the denominator we get:
√9/(4-3i)×(4-3i)/(4-3i)
=[√9(4-3i)]/(16+9)
=[3(4-3i)]/25
=(12-9i)/25

Answer: (12-9i)/25

Answer 2

Solution:

$\frac{\sqrt-9}{[(3-2 i)+(1+5 i)]}$

Remember these complex formulas

1. $\sqrt{-1}=i$

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. $i=\sqrt{-1},i^2=-1, i^3= -i, i^4=1$

→So,$\sqrt{-9}= \sqrt{-1} \times \sqrt{9}= 3 i$

→3 - 2 i + 1 + 5 i= 4 + 3 i

→$\frac{1}{4+3 i}=\frac{4 - 3 i}{(4 + 3 i)(4 - 3 i)}=\frac {4-3 i}{25}$→→Rationalizing the Denominator i.e complex number

→$\frac{\sqrt-9}{[(3-2 i)+(1+5 i)]}$

= $\frac{3 i (4 - 3 i)}{25}=\frac{12 i +9}{25}=\frac{9}{25} +\frac{12 i}{25}$