How do you simplify #(2+5i)/(5+2i)# and write the complex number in standard form?
1 Answer
Jan 8, 2017
Explanation:
The difference of squares identity can be written:
#a^2-b^2 = (a-b)(a+b)#
So we find:
#(5-2i)(5+2i) = 5^2-(2i)^2 = 25+4 = 29#
This provides a way to rationalise the denominator: Multiply both numerator and denominator by the complex conjugate of the denominator thus:
#(2+5i)/(5+2i) = ((5-2i)(2+5i))/((5-2i)(5+2i))#
#color(white)((2+5i)/(5+2i)) = ((5-2i)(2+5i))/29#
#color(white)((2+5i)/(5+2i)) = (overbrace(5(2))^"First"+overbrace(5(5i))^"Outside"+overbrace((-2i)(2))^"Inside"+overbrace((-2i)(5i))^"Last")/29#
#color(white)((2+5i)/(5+2i)) = 1/29(10+25i-4i+10)#
#color(white)((2+5i)/(5+2i)) = 1/29(20+21i)#
#color(white)((2+5i)/(5+2i)) = 20/29+21/29i#
