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Question
State True or False for the following:
Let z1 and z2 be two complex numbers such that |z1 + z2| = |z1| + |z2|, then arg(z1 – z2) = 0.
Options
True
False
Solution
This statement is True.
Explanation:
Let z1 = x1 + y1i and z2 = x2 + y2i
⇒ |z1 + z2| = |z1| + |z2|
⇒ |x1 + y1i + x2 + y2i| = |x1 + y1i| + |x2 + y2i|
⇒ |(x1 + x2) + (y1 + y2)i| = |(x1 + y1i)| + |(x2 + y2i)|
⇒ `sqrt((x_1 + x_2)^2 + (y_1 + y_2)^2) = sqrt(x_1^2 + y_1^2) + sqrt(x_2^2 + y_2^2)`
Squaring both sides, we get
⇒ `(x_1 + x_2)^2 + (y_1 + y_2)^2 = x_1^2 + y_1^2 + x_2^2 + y_2^2 + 2sqrt((x_1^2 + y_1^2)(x_2^2 + y_2^2))`
⇒ `x_1^2 + x_2^2 + 2x_1x_2 + y_1^2 + 2y_1y_2 = x_1^2 + y_1^2 + x_2^2 + y_2^2 + 2sqrt(x_1^2x_2^2 + x_1^2y_2^2 + x_2^2y_1^2 + y_1^2y_2^2)`
⇒ `2x_1x_2 + 2y_1y_2 = 2sqrt(x_1^2x_2^2 + x_1^2y_2^2 + x_2^2y_1^2 + y_1^2y_2^2)`
⇒ `x_1x_2 + y_1y_2 = sqrt(x_1^2x_2^2 + x_1^2y_2^2 + x_2^2y_1^2 + y_1^2y_2^2)`
Again squares on both sides, we get
`x_1^2x_2^2 + y_1^2y_2^2 + 2x_1y_1x_2y_2 = x_1^2x_2^2 + x_1^2y_2^2 + x_2^2y_1^2 + y_1^2y_2^2`
⇒ `2x_1y_1x_2y_2 = x_1^2y_2^2 + x_2^2y_1^2`
⇒ `x_1^2y_2^2 + x_2^2y_1^2 - 2x_1y_1x_2y_2` = 0
⇒ `(x_1y_2 - x_2y_2)^2` = 0
⇒ `x_1y_2 - x_2y_1` = 0
⇒ `x_1y_2 = x_2y_1`
⇒ `x_1/y_1 = x_2/y_2`
⇒ `y_1/x_1 = y_2/x_2`
⇒ arg (z1) = arg (z2)
⇒ arg (z1) – arg (z2) = 0
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