i(z+2z*)=2z+1
i - sq root of -1
z - (a+bi)
z* - conjegate of z - (a-bi)
let z be a+bi
let z* be a-bi
i(a+bi+2(a-bi)=2(a+bi)+1
ia-b+2a-2bi=2a+2bi+1
thats all i can do. please solve for both a and b
2007-03-26 02:12:19 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
i[(a+bi) + 2(a-bi)]= 2(a+bi) + 1
ai -b + 2ai + 2b = 2a+1 + 2bi
b+3ai = (2a+1)+2bi
equating real and imaginary parts :
b = 2a+1
3a = 2b
a = -2, b = -3
Answer is z = -2-3i
2007-03-26 02:25:59 · answer #1 · answered by Nishit V 3 · 1⤊ 0⤋
2) assuming the x+a million is all lower than the sq. root, it can be sqrt 24+a million or sqrt 25 it truly is 5 and -6 circumstances 5 is -30 -a million squared is +a million, circumstances 4 is -4 so it really is -30 - -4 which comes out -30 + 4 or - 26
2016-12-02 20:16:10 · answer #2 · answered by sarro 4 · 0⤊ 0⤋
A=3.5
b= 7
not sure,but I think its the correct answers.
2007-03-26 02:16:15 · answer #3 · answered by Kyle B 4 · 0⤊ 2⤋
a=3.5
b = 7
2007-03-26 02:14:38 · answer #4 · answered by Anonymous · 0⤊ 2⤋