How do you simplify #(2+5i)/(5+2i)# and write the complex number in standard form?

1 Answer
Jan 8, 2017

Answer:

#(2+5i)/(5+2i) = 20/29+21/29i#

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#