Complex no.!!!

z1z2=(1+2i)(3/5 + 4/5 i)=3/5 + 6/5i + 4/5 i - 8/5 = -1+2i

z1/z2= (1+2i)/(3/5 + 4/5 i)=[(1+2i)(3/5 - 4/5 i)]/[(3/5 + 4/5 i)(3/5 - 4/5 i)]
=(3/5 + 6/5i - 4/5 i + 8/5 )/(9/25+16/25)=(11/5+2/5 i)

The distance of z1 to z1z2 and z1z2 to z1/z2 are the same

d1 = |z1-z1z2|=|2|=2
d2 = |z1-z1/z2|=|-6/5-8/5 i| = [(36+64)/25]^0.5=2

the distance of z3 to z1z2 and z1/z2 is also 2

let the z3 = a+bi

d3 = |z3-z1z2|=2
d4 = |z3-z1/z2|=2

|a+bi+1-2i|=2
(a+1)^2+(b-2)^2=4
a^2+2a+1+b^2-4b+4=4
a^2+2a+b^2-4b+1=0 eq1

|a+bi-11/5-2/5 i|=2
(a-11/5)^2+(b-2/5)^2=4

a^2-22/5 a+121/25+b^2-4/5 b+4/25=4
a^2-22/5 a+b^2-4/5 b+1=0 eq2

eq1 - eq2

32/5a-16/5b=0

2a=b
a^2+2a+b^2-4b+1=0
a^2+2a+4a^2-8a+1=0

5a^2-6a+1=0

(5a-1)(a-1)=0
a=1/5 or a=1

Then z3= 1/5(1+2 i) or 1+2i=z1 (reject)

z3= 1/5(1+2 i)

2007-07-12 17:33:57 補充：
Double checking the calculation, pls wait.But the method should work.If z3 = 6/5 + 12/5i |z3|=[(6/5)^2+(12/5)^2]^0.5=6/5 (1^2+2^2)^0.5=6/5 (5)^0.5=6 (5)^0.5

2007-07-12 17:35:08 補充：
|a+bi+1-2i|=2(a+1)^2+(b-2)^2=4a^2+2a+1+b^2-4b+4=4a^2+2a+b^2-4b+5=0 eq1 (Wrong here)|a+bi-11/5-2/5 i|=2(a-11/5)^2+(b-2/5)^2=4a^2-22/5 a+121/25+b^2-4/5 b+4/25=4a^2-22/5 a+b^2-4/5 b+1=0 eq2

2007-07-12 17:38:13 補充：
The above is not correct

2007-07-12 17:41:20 補充：
Would you mind to send me the sketch to my email?

2007-07-12 17:42:06 補充：
I may get misunderstand to you question.