If for complex numbers z1 and z2, arg (z1) – arg (z2) = 0, then show that |z1-z2|=|z1|-|z2|. - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

If for complex numbers z1 and z2, arg (z1) – arg (z2) = 0, then show that `|z_1 - z_2| = |z_1| - |z_2|`.

Advertisement Remove all ads

Solution

Given that for z1 and z2, arg (z1) – arg (z2) = 0.

Let us represent z1 and z2 in polar form.

z1 = r1(cosθ1 + isin θ1) and z2 = r2(cosθ2 + isin θ2)

arg(z1) = θ1 and arg(z2) = θ2

Since arg(z1) – arg(z2) = 0

⇒ θ1 – θ2 = 0

⇒ θ1 = θ2 

Now z1 – z2 = r1(cosθ1 + isinθ1) – r2(cosθ2 + isin θ2)

= r1cosθ1 + ir1sinθ1 – r2cosθ1 – ir2sin θ1   .....`[because theta_1 = theta_2]`

= (r1cosθ1 – r2cosθ1) + i(r1sinθ1 – r2sinθ1)

∴ |z1 – z2| = `sqrt((r_1 costheta_1 - r_2 cos theta_1)^2 + (r_1 sintheta_1 - r_2 sin theta_1)^2`

= `sqrt((r_1^2 cos^2 theta_1 + r_2^2 cos^2 theta_1 - 2r_1r_2 cos^2 theta_1 + r_1^2 sin^2 theta_1 + r_2^2 sin^2 theta_1 - 2r_1r_2 sin^2 theta_1))`

= `sqrt(r_1^2 (cos^2 theta_1 + sin^2 theta_1) + r_2^2 (cos^2 theta_1 + sin^2 theta_1) - 2r_1 r_2 (cos^2 theta_1 + sin^2 theta_1))`

= `sqrt(r_1^2 + r_2^2 - 2r_1r_2)`

= `sqrt((r_1 - r_2)^2`

= r1 – r2

= |z1| – |z2|

Hence, |z1| – |z2| = |z1| – |z2|.

Concept: Argand Plane and Polar Representation
  Is there an error in this question or solution?
Chapter 5: Complex Numbers and Quadratic Equations - Exercise [Page 92]

APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 5 Complex Numbers and Quadratic Equations
Exercise | Q 20 | Page 92

Video TutorialsVIEW ALL [1]

Share
Notifications

View all notifications


      Forgot password?
View in app×