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it is given that complex number z1 and z2 satisfy |Z1|=2 and |z2|=3 if the included angle of the corrossponding vector is 60° then find value of |z1+z2|÷|z1-z2|

it is given that complex number z1 and z2 satisfy |Z1|=2 and |z2|=3 if the included angle of the corrossponding vector is 60° then find value of |z1+z2|÷|z1-z2|

Grade:11

2 Answers

Pratyush Ranjan Roul
19 Points
7 years ago
Ans:   it may be   {133^(1/2)} / {61^(1/2)}  …..............by applying
sunil suthar
13 Points
5 years ago
Using vector law of addition (Of course, we can treat complex numbers as vectors)
 
|z_1 +z_2|^2 = |z_1|^2 + |z_2|^2 +2|z_1||z_2| \cos\theta\\ |z_1 -z_2|^2 = |z_1|^2 + |z_2|^2 -2|z_1||z_2| \cos\theta\\ \text{where}\ \theta\ \text{is the angle between the two complex numbers/vectors}\\ \text{Hence, } \frac{|z_1+z_2|}{|z_1-z_2|}= \sqrt{\frac{|z_1|^2 + |z_2|^2 +2|z_1||z_2| \cos\theta}{|z_1|^2 + |z_2|^2 -2|z_1||z_2| \cos\theta}}\\ = \sqrt\frac{13}{7}\\
 
 
 
\text{Using vector addition} |z_1 +z_2|^2 = |z_1|^2 + |z_2|^2 +2|z_1||z_2| \cos\theta\\ |z_1 -z_2|^2 = |z_1|^2 + |z_2|^2 -2|z_1||z_2| \cos\theta\\ \text{where}\ \theta\ \text{is the angle between the two complex numbers/vectors}\\ \text{Hence, } \frac{|z_1+z_2|}{|z_1-z_2|}= \sqrt{\frac{|z_1|^2 + |z_2|^2 +2|z_1||z_2| \cos\theta}{|z_1|^2 + |z_2|^2 -2|z_1||z_2| \cos\theta}}\\ = \sqrt\frac{19}{7}\\
Regards
SUNIL SUTHAR

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