Question 5 pl Question 5 pl @ II Share with your friends Share 0 Neha Sethi answered this Dear student Let z=a+ib⇒z¯=a-ib⇒z=a2+b2So, a+ib2008=a-ib becomes z2008=z¯Take modulus of both sides z2008=z¯⇒ z2008=z⇒z z2007-1=0⇒z=0 ⇒a2+b2=0⇒a2+b2=0⇒a=0 or b=0or a,b=0,0 ...1or z=1 ...2Given that zz2008=z¯z⇒z2009=z2=1 from 2⇒z2009=1 → This equation has 2009 distinct solutions none of whichis z=0 ..3Results 1 and 3⇒There are 1+2009=2010 distinct solutions⇒There are 2010 distinct pairs that satisfy a+ib2008=a-ib Regards 1 View Full Answer