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October 24th, 2016, 09:14 AM   #1
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Complex numbers question

Hey everyone,

I'm currently making assignments about complex numbers (Calculus, Appendix A7) I've been stuck at some question however, and I can't seem to figure it out, or find any helpful information about it online.

Graph the points z=x+iy that satisfy the following conditions:
a. |z+i|=|z-1|
b. |z+1|≥|z|

Let Re(z) denote the real part of z, and Im(z) denote the imaginary part of z.
Show that the following relations hold for any complex numbers z, z1, and z2.

a. |Re(z)|≤ |z|
b. |z1+z2|≤ |z1|+|z2|

Any help would be very, very much appreciated.

Thank you in advance!

Sincerely,

Karim
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October 24th, 2016, 10:52 AM   #2
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We can interpret "|a- b|" as "the distance between complex number a and complex number b in the complex plane. Saying that |z- a|= |z- b| means that z is equidistant between a and b. That is the line that perpendicularly bisects the line segment from a to b.

We can write any complex number, z, as z= a+ ib for a and b real numbers. In terms of a and b. Re(z)= a, and . Do you see why it is true that tex]|a|\le \sqrt{a^2+ b^2}[/tex]?

Let z1= a+ bi, z2= c +di. Then [tex]|z1+ z2|= \sqrt{(a+ c)^2+ (b+ d)^2}[tex] and .
Look the square of both of those.
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October 24th, 2016, 11:08 AM   #3
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Quote:
Originally Posted by crueltyfate View Post
Let Re(z) denote the real part of z, and Im(z) denote the imaginary part of z.
Show that the following relations hold for any complex numbers z, z1, and z2.

a. |Re(z)|≤ |z|
b. |z1+z2|≤ |z1|+|z2|
$z = Re(z) + i~ Im(z)$

$|z| = \sqrt{(Re(z))^2 + (Im(z))^2} \geq \sqrt{(Re(z))^2} = |Re(z)|$

b) is just the triangle inequality
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