Question

Let z be a complex number, show that \[\sqrt{2}|z| \ge |Re(z)| +|Im(\bar{z})|\]

Answer 1

Perhaps you could use the fact that \(Re(z)=1/2 (z+\bar{z})\) and \(Im(\bar{z})=1/(2i) *(\bar{z}-z)\).

Answer 2

Hmm... It looks more complicated to me.

Answer 3

Well if we square both sides, we get \[(\sqrt2|z|)^2=2z\bar{z}\]and\[Re(z)^2+Im(z)^2+2Re(z)Im(z)\]Rewriting \(z=a+bi\), we get\[2z\bar{z}=2(a^2+b^2)\]and\[a^2+b^2+2ab.\]So we need to show that \(a^2+b^2\ge2ab\).

Answer 4

But then, if we subtract \(2ab\) from both sides, we need to show that \[a^2-2ab+b^2=(a-b)^2\ge 0.\] But since \(a,b\) are real numbers, \(a-b\) is real, so this is true.

Answer 5

Why can't you type slower? /-\
I think I got it. Thanks!!~

Answer 6

You're welcome.

Answer 7

Also, I type math quickly because of lots of \(\LaTeX\) practice.

Answer 8

Practice makes perfect :D

Note: Neglect the following
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Previous attempt
\[z = x+yi\]\[\bar{z} = x-yi\]|Re(z)| = |x| ; |Im(\(\bar{z}\))| = |y|
\[|z| =\sqrt{ x^2 + y^2}\]
To prove:
\[\sqrt{2}|z| \ge |Re(z)| +|Im(\bar{z})|\]
Square both sides
\[2|z|^2 \ge (|Re(z)| +|Im(\bar{z})|)^2\]\[2(x^2+y^2) \ge (x+y)^2\]\[2(x^2+y^2) \ge x^2+2xy+y^2\]\[x^2+y^2 \ge 2xy\]
Not too bad, just not good
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