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