Answer to Question #124661 in Complex Analysis for desmond

Question #124661
(a) 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.
1
Expert's answer
2020-07-02T19:20:07-0400

a) Let "z=x+iy" , then:

"z\\cdot \\overline {z}=(x+iy)(x-iy)=x^2+y^2=|z|^2"

"z+\\overline z=x+iy+x-iy=2x=2Re(z)"

"Re(z)=x=\\sqrt{x^2}\\leq|z|=\\sqrt{x^2+y^2}"


i) Let "z_1=x_1+iy_1,z_2=x_2+iy_2" , then:

"|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=(x_1^2+y_1^2)+(x_2^2+y_2^2)+2(x_1x_2+y_1y_2)="

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


ii)

"|z_1+z_2|=\\sqrt{(x_1^2+y_1^2)+(x_2^2+y_2^2)+2(x_1x_2+y_1y_2)}"

"|z_1|+|z_2|=\\sqrt{x_1^2+y_1^2}+\\sqrt{x_2^2+y_2^2}"

"|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=(x_1^2+y_1^2)+(x_2^2+y_2^2)+2(x_1x_2+y_1y_2)"

"(|z_1|+|z_2|)^2=(x_1^2+y_1^2)+(x_2^2+y_2^2)+2\\sqrt{x_1^2x_2^2+y_1^2y_2^2+x_1^2y_2^2+x_2^2y_1^2}"

Since

"x_1x_2+y_1y_2\\leq\\sqrt{x_1^2x_2^2+y_1^2y_2^2+x_1^2y_2^2+x_2^2y_1^2}"

then

"|z1 + z2| \u2264 |z1| + |z2|"


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