Answer to Question #124528 in Complex Analysis for Amoah Henry

Question #124528
Show that, for any complex number z, zz = |z|
2
, z + z = 2Re(z) and Re(z) ≤ |z|. Hence
show that
i. |z1 + z2|
2 = |z1|
2 + |z2|
2 + 2Re(z1z2),
ii. |z1 + z2| ≤ |z1| + |z2|,
where Re(z) is the real part of z and z the conjugate of z. [26 marks]
(b) If z1 = 1 + 2i, find the set of values of z2 for which
(i) |z1 + z2| = |z1| + |z2| (ii) |z1 + z2| = |z1| − |z2|.
1
Expert's answer
2020-06-30T17:19:40-0400

As "|z|=\\sqrt{x^2+y^2}" where "z=x+iy,\\bar{z}=x-iy," then "z\\sdot\\bar{z}=x^2+y^2=|z|^2" and

"z+\\bar{z}=2x=2Re(z)" where "x=Re(z)" , then "Re(z)\\le |z|"

i. "|z_1+z_2|=(z_1+z_2)(\\bar{z_1}+\\bar{z_2})=z_1\\bar{z_1}+z_1\\bar{z_2}+z_2\\bar{z_1}+z_2\\bar{z_2}="

"=|z_1|^2+|z_2|^2+2Re(z_1z_2)"

ii. Since the sum of two complex numbers can be represented as the third side of a triangle then

"|z_1+z_2|\\le|z_1|+|z_2|"

b) Using the triangle rule

i. If "|z_1+z_2|=|z_1|+|z_2|" then "z_2=k(1+2i)" where "k>0, k\\in R"

ii. If "|z_1+z_2|=|z_1|+|z_2|" then "z_2=-k(1+2i)" where "k>0,k\\in R"


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