Precalculus:Evaluate the expression n a+bi form -4-2i/-3-4i?

Question

I need these by 8pm tonight:
Evaluate the expression (-2-sqrt-9)(3+sqrt-1)

Find all zeros of the polynomial P(x)=x^4-256

I have to work out all of the problems. please help me show details.
thank you. Shagrif

Answer 1

(-4-2i)/(-3-4i) * (-3+4i)/(-3+4i)
= (12 + 6i + 16i + 16i^2)/(9 - 16i^2)
= (12 + 22i - 16)/(9 + 16)
= (-4 + 22i)/25
= -4/25 + 22/25i

(-2-sqrt(-9))(3+sqrt(-1))
sqrt(-9) = 3i (or -3i...but i'm going to assume you only want 3i)
sqrt(-1) = 1i (or -1i, same)
(-2-3i)/(3+i) * (3-i)/(3-i)
= (-6 -9i +2i +2i^2)/(9 - i^2)
= (-6 -2 -7i)/(9 + 1)
= (-8 -7i)/(10)
= -8/10 -7/10i

P(x) is the difference of two squares:
P(x) = (x^2 - 16)(x^2 + 16)
Then x^2 - 16 is also the difference of two squares:
P(x) = (x - 4)(x + 4)(x^2 + 16)
So x = 4, x = -4... (and x = 4i, x = -4i)

Answer 2

Multiply the numerator and denominator by the complex conjugate of the denominator and then divide the product of (-4-2i)(-3-24) by that value.

To evaluate (-2 √-9)(3+√-1), pull out the perfect square (the 9) under the radical in the first termn to get
(-2 +3i)(3+i). Now multiply these together just like any other polynomial (remembering that i² = -1), collect the terms, and you're done.

Remember that (a²-b²) = (a-b)(a+b) and that
(a²+b²) = (a+ib)(a-ib). (HINT: 256 = 16² and 16 = 4²)


Doug

Answer 3

i like those questions :) the important factor to remember is that: i=?-a million i^2=-a million With those, you could resolve it actual. listed decrease than are my steps (btw, dunno the place you're getting the sin and cos from :S): (3 + 4i) ^ 4 =(3 + 4i)(3 + 4i)(3 + 4i)(3 + 4i) =(9+ 12i + 12i +16i ^ 2)(3 + 4i)(3 + 4i) =(-7 + 24i)(3 + 4i)(3 + 4i) =(-21 - 28i + seventy two + 96i ^ 2)(3 + 4i) =(-117 + 44i)(3 + 4i) =-351 - 468i + 132i + 176i ^ 2 =-527 - 336i it is the way I did it and that i've got been given an identical answer as a math application so i've got faith i'm precise, yet others might desire to consistently examine however and determine :) wish this helps!

Answer 4

1)

You need to rationalise the denominator

(-3 - 4i) = -(3+4i)
you mulytiply by (3+4i) both numerator and denominator

numberator becomes -(4+2i)(3+4i) = -(12-8+22i) = -(4+22i)
denominator = -25
number = 4/25 + 22/25i

2) we have (-2-3i)(3+i)
= -6-9i -2i + 3
= -3 - 11i


3)

(x^4-256) = (x^2)^2-16^2= (x^2+16)(x^2-16) using a^2-b^2 = (a+b)(a-b)

x^+16 = 0 gives 2 complex roots 4i and -4i
x^2-16 =0 gives 2 real roots 4 and -4
above 4 are roots