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how to prove \(\large (\overline{z})^n = \overline{z^n}\)
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z is cmplex number over bar refers to conjugate of z
right
certanly true for zconj^1
lets see if we can write one of those ladder +1 proofs
forget what the proof is called induction?
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\(\large (\overline{z})^1 = \overline{z} = \overline{z^1}\) so base n=1 is true
are we trying induction ?
yeah lets try it
okay
assuming \(\large (\overline{z})^n = \overline{z^n}\) is true, we need to show \(\large (\overline{z})^{n+1} = \overline{z^{n+1}}\)
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yeah
i dunno geometrically it kinda just makes sense
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