Prove | z - Re(z) | >= | z - x |

Bronn

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Prove

| z - Re(z) | <= | z - x |

for all real numbers x, draw a sketch to illustrate the result.


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really confused on what to do here. Even after seeing the sketched solution I'm not sure how they got there.

if I let z = x + iy then Re(z) = x and i just get the statement

| iy | >= | iy |

which doesn't really help me much. As far as I can see anyway.

thanks for any help
 
Prove

| z - Re(z) | <= | z - x |

for all real numbers x, draw a sketch to illustrate the result.


-----------------------------------------------------------------------------------------

really confused on what to do here. Even after seeing the sketched solution I'm not sure how they got there.

if I let z = x + iy then Re(z) = x and i just get the statement

| iy | >= | iy |

which doesn't really help me much. As far as I can see anyway.

thanks for any help
Re(z) only equals x if z=x+iy

Try z=a +ib instead.
 
Re(z) only equals x if z=x+iy

Try z=a +ib instead.
I tried that to but I didn't really know what to do with the answer

let z = a+ ib


| a +ib - a| <= | a + ib - x |

| a +ib - a| <= | ( a - x ) + ib |

( ib )^2 <= ( a - x )^2 + ( ib )^2

0 <= a^2 - 2ax + x^2

not sure where to go
 
Last edited:
I tried that to but I didn't really know what to do with the answer

let z = a+ ib


| a +ib - a| <= | a + ib - x | never start a proof with the statement that you are trying to prove - you don't know this is true yet!!

| a +ib - a| <= | ( a - x ) + ib |

( ib )^2 <= ( a - x )^2 + ( ib )^2 (BTW - there should be no i's at this stage)

0 <= a^2 - 2ax + x^2 don't expand it out ... 0<=(a-x)^2 … can you explain why this statement is always true … think of it as (a-x)^2 >=0.

not sure where to go
see comments in red
 
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