Simplify the followings:

A) (a^4 + b^4)/(a^2 +b^2)

B) (2 + 2i)/(3-2i), where i is the imaginary

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- May 17th 2006, 04:20 PMksle82[SOLVED] Hard Algebra Questions
Simplify the followings:

A) (a^4 + b^4)/(a^2 +b^2)

B) (2 + 2i)/(3-2i), where i is the imaginary - May 17th 2006, 04:58 PMtopsquarkQuote:

Originally Posted by**ksle82**

In B) we want to remove the radical in the denominator. So multiply top and bottom by the complex conjugate of the denominator:

$\displaystyle \frac{2+2i}{3-2i} \cdot \frac{3+2i}{3+2i}$

Since $\displaystyle i^2=-1$:

$\displaystyle =\frac{6+4i+6i+4i^2}{9+6i-6i-4i^2}=\frac{6+10i-4}{9+4}$

$\displaystyle =\frac{2+10i}{13}$

-Dan - May 17th 2006, 05:27 PMticbolQuote:

Originally Posted by**ksle82**

A) (a^4 +b^4) / (a^2 +b^2) --------(1)

A.1) By long division, = (a^2 -b^2) +(2b^4)/(a^2 +b^2) -----------------------answer.

A.2) Get a factor (a^2 +b^2) from the numerator (a^4 +b^4):

Squaring the denominator,

(a^2 +b^2)^2 = a^4 +2(a^2)(b^2) +b^4

(a^2 +b^2)^2 = (a^4 +b^4) +2(a^2)(b^2)

(a^4 +b^4) = (a^2 +b^2)^2 -2(a^2)(b^2) ------(i)

Substitute that into (1),

(a^4 +b^4) / (a^2 +b^2) --------(1)

= [(a^2 +b^2)^2 -2(a^2)(b^2)] / (a^2 +b^2)

= (a^2 +b^2) -2(a^2)(b^2)/(a^2 +b^2) --------------answer.

=============================

B) (2 +2i) / (3 -2i) -----------------------(2)

Make the denominator a real number, muliply both numerator and denominator by the conjugate of the denominator, that is by (3 +2i):

= [(2 +2i)(3 +2i)] / [(3 -2i)(3 +2i)]

= [6 +4i +6i +4i^2] / [9 -4i^2]

Since i^2 = -1, then,

= [2 +10i] / [13]

= (1/13)(2 +10i)

= (2/13)(1 +5i) -----------answer.