The equation of the line joining the complex numbers -5+4i and 7+2i can be expressed in the form
az + b \overline{z} = 38
for some complex numbers a and b. Find the product ab.
The equation of the line joining the complex numbers
−5+4i and 7+2i can be expressed in the form
az+b¯z=38
for some complex numbers a and b. Find the product ab.
Let z1=−5+4i and ¯z1=−5−4iLet z2=7+2i and ¯z2=7−2iLet az1+b¯z1=38Let az2+b¯z2=38
az2+b¯z2=38az2=38−b¯z2a=38−b¯z2z2az1+b¯z1=3838−b¯z2z2z1+b¯z1=38|∗z2(38−b¯z2)z1+b¯z1z2=38z238z1−b¯z2z1+b¯z1z2=38z2b(¯z1z2−¯z2z1)=38(z2−z1)b=38(z2−z1)(¯z1z2−¯z2z1)¯z1z2−¯z2z1=(−5−4i)(7+2i)−(7−2i)(−5+4i)=−76i
b=38(z2−z1)¯z1z2−¯z2z1=38(7+2i−(−5+4i))−76i=38(7+2i+5−4i)−76i=7+2i+5−4i−2i=12−2i−2i=−6+ii=−6i+1=−6ii2+1=−6i−1+1=6i+1b=1+6i
az2+b¯z2=38az2+38(z2−z1)(¯z1z2−¯z2z1)¯z2=38az2=38−38(z2−z1)¯z2(¯z1z2−¯z2z1)|:z2a=38z2−38(z2−z1)¯z2(¯z1z2−¯z2z1)z2a=38(¯z1z2−¯z2z1)−38(z2−z1)¯z2z2(¯z1z2−¯z2z1)a=38(¯z1z2−¯z2z1−(z2−z1)¯z2)z2(¯z1z2−¯z2z1)a=38(¯z1z2−¯z2z1−z2¯z2+z1¯z2)z2(¯z1z2−¯z2z1)a=38(¯z1z2−z2¯z2)z2(¯z1z2−¯z2z1)a=38z2(¯z1−¯z2)z2(¯z1z2−¯z2z1)a=38(¯z1−¯z2)(¯z1z2−¯z2z1)
a=38(¯z1−¯z2)(¯z1z2−¯z2z1)=38(−5−4i−(7−2i))−76i=38(−7+2i−5−4i)−76i=−7+2i−5−4i−2i=−12−2i−2i=6+ii=6i+1=6ii2+1=6i−1+1=−6i+1a=1−6i
ab=(1+6i)(1−6i)ab=1+36ab=37