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.

2 Answers
Cesareo R.
Mar 14, 2017

#(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#

Steve M · Shwetank Mauria
Mar 14, 2017

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#

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