Imaginary part of z

Dec 2015
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Earth
Find y such that \(\displaystyle Im(z)=0\).

\(\displaystyle z=\dfrac{1}{1+yi}-\dfrac{1}{1+y^2 i }\).
 

mathman

Forum Staff
May 2007
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y=0 and y=1 are solutions. With a little algebra, it can be shown these are the only solutions.
 
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skipjack

Forum Staff
Dec 2006
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Is $y$ required to be real?
 

skipjack

Forum Staff
Dec 2006
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W|A finds a solution for which $y$ isn't real.
 
Dec 2015
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169
Earth
\(\displaystyle Im(z)=[\dfrac{i}{i-y} - \dfrac{i}{i-y^2 }]_{i} ' =0 \; \) ; \(\displaystyle \; (y-y^2 )(y^3 -1)=0\).
\(\displaystyle y^2 =y\) has real solutions \(\displaystyle y=0 \vee y=\pm 1\).
\(\displaystyle y^3=1\) has comlex solutions \(\displaystyle y=\dfrac{1}{2} \pm i\dfrac{\sqrt{3}}{2}\).
 
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