If a = operatorname{Im}left(frac{1+z^2}{2iz}r | vedclass

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(C) Given that $|z|=1$,we can write $z = x + iy$ where $x^2 + y^2 = 1$. Then,$\frac{1+z^2}{2iz} = \frac{1}{2i} \left(\frac{1}{z} + z\right)$. Since $|

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$-\operatorname{Re}(z)$

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(C) Given that $|z|=1$,we can write $z = x + iy$ where $x^2 + y^2 = 1$. Then,$\frac{1+z^2}{2iz} = \frac{1}{2i} \left(\frac{1}{z} + z\right)$. Since $|z|=1$,we have $\frac{1}{z} = \bar{z} = x - iy$. Substituting this,we get $\frac{1}{2i} (x - iy + x + iy) = \frac{1}{2i} (2x) = \frac{x}{i} = -ix$. Thus,$\frac{1+z^2}{2iz} = -ix$. Taking the imaginary part,$a = \operatorname{Im}(-ix) = -x$. Since $x = \operatorname{Re}(z)$,we have $a = -\operatorname{Re}(z)$.

If $a = \operatorname{Im}\left(\frac{1+z^2}{2iz}\right)$ and $z$ is any non-zero complex number such that $|z|=1$,then $a=$

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