Complex numbers problem
- Thread starter Sophdof1
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pka
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Sophdof1, many of us do not use a phone, pad, or a laptop so when you post am image which is sideways it makes it almost impossible to read on a desktop.
That said, for a complex number \(\displaystyle z\) it is true that \(\displaystyle \|z^2\|=\|z\|^2\) so \(\displaystyle \|(z-2)^2\|=\|z-2\|^2\).
Now please have a look HERE.
What you've done so far seems fine to me. As for a next step, maybe try using the most generic form of a complex number there is. Let \(z = a + bi\):
\(\displaystyle |(a + bi - 2)^2| < |(a + bi - 1)^2|\)
\(\displaystyle |[(a - 2) + bi]^2| < |[(a - 1) + bi]^2|\)
Where does this lead you?
\(\displaystyle |(a + bi - 2)^2| < |(a + bi - 1)^2|\)
\(\displaystyle |[(a - 2) + bi]^2| < |[(a - 1) + bi]^2|\)
Where does this lead you?
Dr.Peterson
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Good work so far. I would continue by combining pka and ksdhart2's ideas:
||z-2||^2 < ||z-1||^2
||(a-2) + bi||^2 < ||(a-1) + bi||^2
Now use the formula for ||x + yi||^2 on each side. You'll end up with a quadratic inequality - well, actually much simpler than that!
||z-2||^2 < ||z-1||^2
||(a-2) + bi||^2 < ||(a-1) + bi||^2
Now use the formula for ||x + yi||^2 on each side. You'll end up with a quadratic inequality - well, actually much simpler than that!