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March 29, 2024, 06:33:24 pm

Author Topic: Complex number question  (Read 1170 times)  Share 

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Jimbo123

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Complex number question
« on: October 16, 2019, 10:21:36 pm »
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Hi all

Have a quick complex number question

If both z+w and zw are real, prove that either z=wbar or Im(z) = Im(w) = 0

Thanks

RuiAce

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Re: Complex number question
« Reply #1 on: October 16, 2019, 10:54:48 pm »
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Hi all

Have a quick complex number question

If both z+w and zw are real, prove that either z=wbar or Im(z) = Im(w) = 0

Thanks
\[ \text{Let }z=x+iy\text{ and }w=u+iv. \]
\[ \DeclareMathOperator{\Im}{Im} \text{If }z+w\text{ and }zw\text{ are real, then } \boxed{\Im(z+w) = 0}\text{ and }\boxed{\Im(zw) = 0}.\]
\[ \text{Since }z+w = (x+u) + i(y+v),\text{ this would give us }y+v = 0 \implies \boxed{v=-y}.\\ \text{Since }zw = (xu - yv) + i(yu + xv)\text{, this would give us }\boxed{yu+xv = 0}. \]
\[ \text{Hence upon subbing,}\\ \begin{align*} yu-xy &= 0\\ y(u-x) &= 0 \end{align*}\\ \text{Therefore }y=0\text{ or }u=x.\]
\[ \text{In the case }y=0\text{, subbing back also gives }v=0.\\ \DeclareMathOperator{\Im}{Im}\text{But by definition of }z\text{ and }w, \, \Im(z) = y\text{ and }\Im(w) = v.\\ \text{Hence this case correctly yields }\boxed{\Im(z)=\Im(w) = 0} \]
\[ \text{In the case }u=x\text{, we note that this effectively gives the conditions}\\ \DeclareMathOperator{\Re}{Re}\begin{align*} \Re(w) &= \Re(z),\\ \Im(z) &= -\Im(w). \end{align*}\\ \text{These are, in fact, the exact conditions required to satisfy }\boxed{z=\overline{w}}. \]
Alternatively for the second case, subbing back gives \(w = x-iy\), which makes it more obvious that \(w = \overline{z}\).
« Last Edit: October 16, 2019, 10:58:31 pm by RuiAce »