Circling complex!

Algebra Level 4

Let complex numbers $\alpha$ and $\frac { 1 }{ \bar { \alpha } }$ lies on circles ${ \left( x-{ x }_{ 0 } \right) }^{ 2 }+{ \left( y-{ y }_{ 0 } \right) }^{ 2 }={ r }^{ 2 }$ and ${ \left( x-{ x }_{ 0 } \right) }^{ 2 }+{ \left( y-{ y }_{ 0 } \right) }^{ 2 }=4{ r }^{ 2 }$ respectively.

If ${ z }_{ 0 }={ x }_{ 0 }+i{ y }_{ 0 }$ satisfies the equation $2{ \left| { z }_{ 0 } \right| }^{ 2 }={ r }^{ 2 }+2$ then find the value of $\left| \alpha \right|$

The answer is 0.378.

2 solutions

Mudit Bansal
Feb 20, 2015

Given: $\left| z-{ z }_{ 0 } \right| =r\quad and\quad \left| z-{ z }_{ 0 } \right| =2r$ Now, $\alpha$ and $\frac { 1 }{ \bar { \alpha } }$ satisfies these equations in respective order.Therefore, on substituting and squaring we get: $\left( \alpha -{ z }_{ 0 } \right) \left( \bar { \alpha } -\bar { { z }_{ 0 } } \right) ={ r }^{ 2 }\quad and\quad \left( \frac { 1 }{ \bar { \alpha } } -{ z }_{ 0 } \right) \left( \frac { 1 }{ \alpha } -\bar { { z }_{ 0 } } \right) =4{ r }^{ 2 }$ On subtracting the two equations we get: ${ \left| \alpha \right| }^{ 2 }-1+{ \left| { z }_{ 0 } \right| }^{ 2 }\left( 1-{ \left| \alpha \right| }^{ 2 } \right) ={ r }^{ 2 }\left( 1-{ 4\left| \alpha \right| }^{ 2 } \right)$ Now given that: ${ \left| { z }_{ 0 } \right| }^{ 2 }=\frac { { r }^{ 2 }+2 }{ 2 }$ Hence,on substituting and simplifying we get: ${ \left| \alpha \right| }^{ 2 }=\frac { 1 }{ 7 } \\ \left| \alpha \right| =\frac { 1 }{ \sqrt { 7 } } \simeq \boxed { 0.378 }$

Srivats Anshumaali
Apr 30, 2015

i think we can apply a little trigonometry notice alpha and 1/alpha (conjugate) make same angles hence a part of same line and then use cosine formula

Can you elaborate on this?

Pi Han Goh - 5 years, 6 months ago

mod(alpha-z)^2= mod(alpha)^2+mod(z)^2 -2* mod(alpha)mod(z)cos(theta) A,
where theta is the angle between alpha and z. The angle will be the same between 1/alpha(bar) and z. Similarly you can write mod(1/alpha(bar)-z)^2= mod(1/alpha(bar))^2+mod(z)^2-2 mod(1/alpha(bar))mod(z)cos(theta) B. Now from the equations A and B eliminate cos theta and use the fact that mod(alpha(bar))=mod(alpha) and 2 mod(z)=r^2+2 you should get after rearranging (7 *mod(alpha)^2 -1)(mod (z)^2 -1)=0. If mod(z)^2=1 then r=0 it will be a degenerate case so mod(alpha)=1/sqrt7 PS: The computation part is a bit lengthy but if we attempt it carefully the solution is shorter and creative hope that helps, sorry I cannot draw a diagram to show you but it is not tough to visualize the geometric picture

Srivats Anshumaali - 5 years, 6 months ago

Circling complex!

The answer is 0.378.

2 solutions

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