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BartSMP [9]
3 years ago
13

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

Mathematics
2 answers:
jeka943 years ago
8 0
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
anzhelika [568]3 years ago
7 0

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}




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