If {z_1}{rm{ and }}{z_2} be complex numbers s | vedclass

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Question

(a)Let ${z_1} = a + ib = (a,b)$and ${z_2} = c - id = (c, - d)$
Where $a > 0$and $d > 0$ ......$(i)$
Then $|{z_1}| = |{z_2}|$ $&rArr;$&nbs

Vedclass · Accepted Answer

Purely imaginary

Vedclass · Answer

(a)Let ${z_1} = a + ib = (a,b)$and ${z_2} = c - id = (c, - d)$
Where $a > 0$and $d > 0$ ......$(i)$
Then $|{z_1}| = |{z_2}|$ $&rArr;$  ${a^2} + {b^2} = {c^2} + {d^2}$
Now $\frac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}} = \frac{{(a + ib) + (c - id)}}{{(a + ib) - (c - id)}}$
$ = \frac{{[(a + c) + i(b - d)][(a - c) - i(b + d)]}}{{[(a - c) + i(b + d)][(a - c) - i(b + d)]}}$
$ = \frac{{({a^2} + {b^2}) - ({c^2} + {d^2}) - 2(ad + bc)i}}{{{a^2} + {c^2} - 2ac + {b^2} + {d^2} + 2bd}}$ $\frac{{ - (ad + bc)i}}{{{a^2} + {b^2} - ac + bd}}$[using $(i)$]

$	herefore $$\frac{{({z_1} + {z_2})}}{{({z_1} - {z_2})}}$ is purely imaginary.

However if $ad + bc = 0$, then $\frac{{({z_1} + {z_2})}}{{({z_1} - {z_2})}}$ will be equal to zero. According to the conditions of the equation, we can have $ad + bc = 0$
Trick : Assume any two complex numbers satisfying both conditions i.e., ${z_1} 
e {z_2}$and $|{z_1}|\, = \,|{z_2}|$
Let ${z_1} = 2 + i,{z_2} = 1 - 2i,$

$	herefore \,\,\frac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}} = \frac{{3 - i}}{{1 + 3i}} = - i$
Hence the result.

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

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