Saved Bookmarks
| 1. |
if z lies on |z| = 1, then \(\rm \frac 2 z\) lies on1. a circle2. an ellipse3. a straight line 4. None of these |
|
Answer» Correct Answer - Option 1 : a circle Concept: Let z = x + iy be a complex number. Modulus of z = \(\left| {\rm{z}} \right| = \sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = \sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{\;Im\;}}{{\left( {\rm{z}} \right)}^2}} \) \(\left| {\frac{{{{\rm{z}}_1}}}{{{{\rm{z}}_2}}}} \right| = {\rm{\;}}\frac{{\left| {{{\rm{z}}_1}} \right|}}{{\left| {{{\rm{z}}_2}} \right|}},{\rm{\;}}\left| {{{\rm{z}}_2}} \right| \ne 0\)
Equation of a circle
Calculation: Given: |z| = 1 Let z' = \(\rm \frac 2 z\) Taking the mod of both sides, we get ⇒ |z'| = \(\rm \left| \frac 2 z \right|\) ⇒ |z'| = \(\rm \frac {|2|}{| z|}\) ⇒ |z'| = \(\frac 21\) (∵ |z| = 1) ⇒ |z'| = 2 it will be a circle of radius 2 units and with centre as origin. |
|