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Question : 83 of 120
Marks:
+1,
-0
Solution:
Given,
|z1+z2|=|z1|+|z2| Let
‌‌z1=x1+iy1 z2=x2+iy2
|(x1+x2)+i(y1+y2)|=|x1+iy1|+|x2+iy2|
√(x1+x2)2+(y1+y2)2=√x12+y12+√x22+y22
Squaring on both sides
x12+x22+2x1x2+y12+y22+2y1y2
=x12+y12+x22+y22+2√(x12+y12)(x22+y22)
⇒‌‌2(x1x2+y1y2)=2√(x12+y12)(x22+y22)
⇒‌‌(x1x2+y1y2)2=(x12+y12)(x22+y22)
⇒‌‌x12x22+y12y22+2x1x2y1y2 =x12x22+x22y12+x12y22+y12y22 ⇒‌‌2x1x2y1y2=x22y12+x12y22 ⇒‌‌(−x1y2+y1x2)2=0 ⇒‌‌x2y1−y2x1=0......(i)
‌=‌ =‌| (x1+iy1)(x2−iy2) |
| (x22−y22) |
=‌+i‌ arg(‌)=tan−1(‌) =tan−1(‌) =tan−1(0)=0‌‌[ from
Eq (i)]
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