If #(a+bi)=((4+4i)^7)/((4-4i)^8)# what is the value of #a#?

1 Answer
Feb 12, 2016

#a=1/8#

Explanation:

Observations (to be used later):
#color(white)("XXX")(1+i)/(1-i) = ((1+i) * (1+i))/((1-i) * (1+i)) = (2i)/2 = i#

#color(white)("XXX")i^7 = i^3 = -i#

Evaluating:
#(4+4i)^7/(4-4i)^8#

#color(white)("XXX")= (4^7*(1+i)^7)/(4^8*(1-i)^8)#

#color(white)("XXX")=1/4 * ((1+i)/(1-i))^7*1/(1-i)#

#color(white)("XXX")=1/4*i^7*1/(1-i)#

#color(white)("XXX")=1/4*i/(i-1)#

#color(white)("XXX")=1/4*(i*(i+1))/((i-1)*(i+1))#

#color(white)("XXX")= 1/4* ((-1+i))/((-2))#

#color(white)("XXX")=1/8-1/8i#

If #(a+bi) = 1/8-1/8i rarr a=1/8#