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(A) Let ${z_1} = a + ib$ and ${z_2} = c - id$,where $a > 0$ and $d > 0$. Given $|{z_1}| = |{z_2}|$,we have ${a^2} + {b^2} = {c^2} + {d^2}$. Let $w = \

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Purely imaginary

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(A) Let ${z_1} = a + ib$ and ${z_2} = c - id$,where $a > 0$ and $d > 0$. Given $|{z_1}| = |{z_2}|$,we have ${a^2} + {b^2} = {c^2} + {d^2}$. Let $w = \frac{{z_1 + z_2}}{{z_1 - z_2}}$. Then $\bar{w} = \frac{{\bar{z_1} + \bar{z_2}}}{{\bar{z_1} - \bar{z_2}}}$. Since $|{z_1}| = |{z_2}| = r$,we have $\bar{z_1} = \frac{r^2}{z_1}$ and $\bar{z_2} = \frac{r^2}{z_2}$. Substituting these,$\bar{w} = \frac{{\frac{r^2}{z_1} + \frac{r^2}{z_2}}}{{\frac{r^2}{z_1} - \frac{r^2}{z_2}}} = \frac{{z_2 + z_1}}{{z_2 - z_1}} = -\frac{{z_1 + z_2}}{{z_1 - z_2}} = -w$. Since $\bar{w} = -w$,the complex number $w$ is purely imaginary. For example,let ${z_1} = 2 + i$ and ${z_2} = 1 - 2i$. Then $\frac{{z_1 + z_2}}{{z_1 - z_2}} = \frac{{3 - i}}{{1 + 3i}} = \frac{{(3 - i)(1 - 3i)}}{{1^2 + 3^2}} = \frac{{3 - 9i - i - 3}}{{10}} = \frac{{-10i}}{{10}} = -i$,which is purely imaginary.

If ${z_1}$ and ${z_2}$ are complex numbers such that ${z_1} \neq {z_2}$ and $|{z_1}| = |{z_2}|$. If ${z_1}$ has a positive real part and ${z_2}$ has a negative imaginary part,then $\frac{{z_1 + z_2}}{{z_1 - z_2}}$ may be

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