Algebra> it is given that complex number z1 and 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|

Priyank Kushwah , 8 Years ago

Grade 11

2 Answers

Pratyush Ranjan Roul

Last Activity: 8 Years ago

Ans: it may be {133^(1/2)} / {61^(1/2)} …..............by applying

sunil suthar

Last Activity: 6 Years ago

Using vector law of addition (Of course, we can treat complex numbers as vectors)

$|z_1 +z_2|^2 = |z_1|^2 + |z_2|^2 +2|z_1||z_2| \cos\theta\\ |z_1 -z_2|^2 = |z_1|^2 + |z_2|^2 -2|z_1||z_2| \cos\theta\\ \text{where}\ \theta\ \text{is the angle between the two complex numbers/vectors}\\ \text{Hence, } \frac{|z_1+z_2|}{|z_1-z_2|}= \sqrt{\frac{|z_1|^2 + |z_2|^2 +2|z_1||z_2| \cos\theta}{|z_1|^2 + |z_2|^2 -2|z_1||z_2| \cos\theta}}\\ = \sqrt\frac{13}{7}\\$

$\text{Using vector addition} |z_1 +z_2|^2 = |z_1|^2 + |z_2|^2 +2|z_1||z_2| \cos\theta\\ |z_1 -z_2|^2 = |z_1|^2 + |z_2|^2 -2|z_1||z_2| \cos\theta\\ \text{where}\ \theta\ \text{is the angle between the two complex numbers/vectors}\\ \text{Hence, } \frac{|z_1+z_2|}{|z_1-z_2|}= \sqrt{\frac{|z_1|^2 + |z_2|^2 +2|z_1||z_2| \cos\theta}{|z_1|^2 + |z_2|^2 -2|z_1||z_2| \cos\theta}}\\ = \sqrt\frac{19}{7}\\$