MCQ
$|{z_1} + {z_2}|\, = \,|{z_1}| + |{z_2}|$ is possible if
- A${z_2} = {\overline z _1}$
- B${z_2} = \frac{1}{{{z_1}}}$
- ✓$arg\,({z_1}) = arg ({z_2})$
- D$|{z_1}|\, = \,|{z_2}|$
✓
Answer
Correct option: C.
$arg\,({z_1}) = arg ({z_2})$
c
(c) $|{z_1} + {z_2}|\, = \,|{z_1}| + |{z_2}|$
$|{z_1} + {z_2}{|^2}\, = \,|{z_1}{|^2} + |{z_2}{|^2} + 2|{z_1}||{z_2}|$
==> $|{z_1}{|^2} + |{z_2}{|^2} + 2{\mathop{\rm Re}\nolimits} |{z_1}{\bar z_2}|$
$ = \,|{z_1}{|^2} + |{z_2}{|^2} + 2|{z_1}||{z_2}|$
==> $2{\mathop{\rm Re}\nolimits} |{z_1}{\bar z_2}| = 2|{z_1}||{z_2}|$
==> $2|{z_1}||{\bar z_2}|\cos ({\theta _1} - {\theta _2}) = 2|{z_1}||{z_2}|$
==> $\cos ({\theta _1} - {\theta _2}) = 1$or ${\theta _1} - {\theta _2} = 0$
$arg({z_1}) = arg({z_2})$
(c) $|{z_1} + {z_2}|\, = \,|{z_1}| + |{z_2}|$
$|{z_1} + {z_2}{|^2}\, = \,|{z_1}{|^2} + |{z_2}{|^2} + 2|{z_1}||{z_2}|$
==> $|{z_1}{|^2} + |{z_2}{|^2} + 2{\mathop{\rm Re}\nolimits} |{z_1}{\bar z_2}|$
$ = \,|{z_1}{|^2} + |{z_2}{|^2} + 2|{z_1}||{z_2}|$
==> $2{\mathop{\rm Re}\nolimits} |{z_1}{\bar z_2}| = 2|{z_1}||{z_2}|$
==> $2|{z_1}||{\bar z_2}|\cos ({\theta _1} - {\theta _2}) = 2|{z_1}||{z_2}|$
==> $\cos ({\theta _1} - {\theta _2}) = 1$or ${\theta _1} - {\theta _2} = 0$
$arg({z_1}) = arg({z_2})$
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