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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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.Quote:
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:Quote:
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.
