Question

Finding the locus of a complex number?

`z` is a variable complex number such that `|z|=1` and `u=3z-1/z`. Show that the locus of the point in the Argand plane representing `u` is an ellipse and find the equation of the ellipse.

Answer 1

`(X/2)^2+(Y/4)^2=1`

Explanation:

We have `z bar z= abs z^2=x^2+y^2=1` This can be described also as

`z=e^(it)` for `e^(it)=cos(t)+isin(t) = x(t) + i y(t)`

Now

`u=3e^(it)-e^(-(it))=3(cos(t)+isin(t))-(cos(-t)+isin(-t))`

or

`u=3cos(t)-cos(t)+i(3sin(t)+sin(t)) = X+iY`

or

`{(3cos(t)-cos(t)=2cos(t)=X),(3sin(t)+sin(t)=4sin(t)=Y):}`

so the ellipse equation is obtained as

`cos^2(t)+sin^2(t)=(X/2)^2+(Y/4)^2=1`

Answer 2

Loci is that of an ellipse with equation

`(x/2)^2 + (y/4)^2 =1`

Explanation:

Let `u=x+yi`, and `z=a+bi`; then

From `|z|=1` we have:

`a^2 + b^2 = 1 \ \ \ \ .....[1]`

From `u=3z-1/z` we have:

`x+yi = 3(a+bi) - 1/(a+bi)`
`" " = 3a+3bi - 1/(a+bi) * (a-bi)/(a-bi)`
`" " = 3a+3bi - (a-bi)/(a^2-(bi)^2)`
`" " = 3a+3bi - (a-bi)/(a^2+b^2)`
`" " = 3a+3bi - (a-bi) \ \ \ ` (from [1]) # " " = 3a+3bi - a+bi " " = 2a+4bi #

And so:

`x=2a => a=x/2`
`y=4b => b=y/4`

Squaring and adding we get:

`a^2 + b^2 = (x/2)^2 + (y/4)^2`

`:. (x/2)^2 + (y/4)^2 =1` (from [1])

Which is the equation f an ellipse with semi-minor axis `2` and semi-major axis `4`