Simplify the followings:
A) (a^4 + b^4)/(a^2 +b^2)
B) (2 + 2i)/(3-2i), where i is the imaginary
For the way most would want the answer, A) is already as simple as it is going to get.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
Here are some ways: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.