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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.

 Jun 5, 2022
 #1
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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=54iLet z2=7+2i and ¯z2=72iLet az1+b¯z1=38Let az2+b¯z2=38

 

az2+b¯z2=38az2=38b¯z2a=38b¯z2z2az1+b¯z1=3838b¯z2z2z1+b¯z1=38|z2(38b¯z2)z1+b¯z1z2=38z238z1b¯z2z1+b¯z1z2=38z2b(¯z1z2¯z2z1)=38(z2z1)b=38(z2z1)(¯z1z2¯z2z1)¯z1z2¯z2z1=(54i)(7+2i)(72i)(5+4i)=76i

 

b=38(z2z1)¯z1z2¯z2z1=38(7+2i(5+4i))76i=38(7+2i+54i)76i=7+2i+54i2i=122i2i=6+ii=6i+1=6ii2+1=6i1+1=6i+1b=1+6i

 

az2+b¯z2=38az2+38(z2z1)(¯z1z2¯z2z1)¯z2=38az2=3838(z2z1)¯z2(¯z1z2¯z2z1)|:z2a=38z238(z2z1)¯z2(¯z1z2¯z2z1)z2a=38(¯z1z2¯z2z1)38(z2z1)¯z2z2(¯z1z2¯z2z1)a=38(¯z1z2¯z2z1(z2z1)¯z2)z2(¯z1z2¯z2z1)a=38(¯z1z2¯z2z1z2¯z2+z1¯z2)z2(¯z1z2¯z2z1)a=38(¯z1z2z2¯z2)z2(¯z1z2¯z2z1)a=38z2(¯z1¯z2)z2(¯z1z2¯z2z1)a=38(¯z1¯z2)(¯z1z2¯z2z1)

 

a=38(¯z1¯z2)(¯z1z2¯z2z1)=38(54i(72i))76i=38(7+2i54i)76i=7+2i54i2i=122i2i=6+ii=6i+1=6ii2+1=6i1+1=6i+1a=16i

 

ab=(1+6i)(16i)ab=1+36ab=37

 

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 Jun 6, 2022

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