Evaluate the expression and write your result in the form a + bi

1) 2i(1/2 - i)


2) 5 - i/3 + 4i


3) (1/1 + i) - (1/1 - i)



Factor the polynomial completely, and find all of its zeros. State the multiplicity of every zero

1) Q(x) = x^4 + 10x^2 +25


Find a polynomial with integer coefficients that satisfies the given conditions

1) P has a degree 2 and zeros 1 + i(sqrt 2) and 1 - i(sqrt 2)


These are the only ones I have left to do. Any help would be appreciated.

This post was edited by JoeyMorgan619 on Mar 26 2014 10:10pm

1) 2) 3) , do those normally , think of i as a variable (like x)
leave in the form of a+bi , a and b are constants
(5+6i) + (6+3i) = (11+9i)
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1)
5*5=25 ,5+5=10 is a start

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(x-a)(x-b)

This post was edited by saber_x3 on Mar 26 2014 10:43pm

still can't get this one...can anyone help?

1/(1+i) - 1/(1-i)

In order to subtract one term from the other, they need to have a common denominator. So you need to change both terms so that they have the same denominator. This can be done by multiplying each one by a fraction that is actually equal to 1, because you can always multiply something by 1 and not change its value.

1(1-i)/(1+i)(1-i) - 1(1+i)/(1-i)(1+i)

Now that they have the same denominator you can subtract.

[(1-i)-(1+i)]/(1-i)(1+i)
-2i/(1-i)(1+i)

Distribute the denominator.

-2i/(1 + i - i - i^2)
-2i/(1-i^2)

i^2 is equal to -1.

-2i/[1-(-1)]
-2i/2

Simplify.

-i

first step multiply each part with denominator/denominator of the other part, ie

(1-i)/[(1+i)(1-i)] - (1+i)/[(1-i)(1+i)] now notice that (1+i)(1-i)=1+1=2 and you have

(1-i)/2 - (1+i)/2 = [1-i-1-i]/2 =[-2i] / 2 = -i

postscript: my typing was to slow 'katharsis' gave you the answer with the same approach but differently ordered steps

This post was edited by brmv on Mar 29 2014 08:38pm