How to solve the equation #z^4=ibarz# ? Where z is a complex number.

2 Answers
Mar 5, 2017

Answer:

#S = { 0, e^((ipi)/10) } #

Explanation:

Let's think of these complex numbers in exponential forms!

#z# is a complex number of phase, or angle #theta#, and its conjugate has the angle #-theta#, we can see that by algebra if we use certain trig properties (the cosine is even and the sine is odd):

#z = re^(itheta) = cos(theta) + i*sin(theta) = a + bi#
#barz = re^(-itheta) = cos(-theta) + isin(-theta) = cos(theta) -isin(theta) = a -bi#

So, rewriting this in terms of angles, we have:

#re^(i*4theta) = i*re^(-itheta)#

Remembering that #i# also has an exponential form, #e^(pi/2*i)#

#re^(i*4theta) = re^(i(pi/2-theta))#
#i*4theta = i(pi/2 - theta)#
#4theta = pi/2 - theta#
#5theta = pi/2#
#theta = pi/10#

Now to solve for the radius, we know that:

#z^4 = ibarz#

By taking the absolute value of both sides we have

#|z^4| = |ibarz|#

Which we can split

#|z|^4 = |i||barz|#

But #|z| = |barz|# and it's defined as the radius we want to discover, so

#r^4 = |i|*r#

#|i| = 1#, so simplifying we have

#r^4 = r#

There are only two real numbers (remember that the absolute value / radius is always real) that can satisfy this equation

#r = 1# or #r = 0#

Mar 5, 2017

Answer:

# z=0,i #
# z = +-1/2sqrt(1/2(5+sqrt(5))) + 1/4(sqrt(5)-1)i#
# z = +-1/2sqrt(1/2(5-sqrt(5))) - 1/4(sqrt(5)+1)i#

Explanation:

We have:

# z^4 = i barz #,

Where #barz # is the complex conjugate of #z#

If we use the exponential form of a complex number then we can represent the solution by;

# z= re^(itheta) \ \ \ (=rcostheta+irsintheta)#

Then the complex conjugate will be:

# z= re^(-itheta) #

Whenever dealing with complex equation such as this it is essential to remember that the complex exponential has a period of #2pi#, and thus we can modify the conjugate, and substitute into the equation to solve to get;

# {re^(itheta)}^4 = i(re^(-itheta)) #
# {re^(itheta)}^4 = i(re^(i(-theta+2npi))) # where # n in NN#

(Equally we could have incorporated this periodicity into the RHS exponent, it really does not matter providing it is incorporated somewhere). Replacing #i# by its complex exponential form #i=e^(ipi/2)# we get:

# {re^(itheta)}^4 = e^(ipi/2)(re^(i(-theta+2npi))) #
# :. r^4e^(4itheta) = re^(i(pi/2-theta+2npi)) #

Equating real and imaginary terms (or the moduli and arguments) we have:

# Re\ : \ \ r^4=r #
# Im: \ \ 4theta = pi/2-theta +2npi#

From the first equation we get:

# r^4-r = 0#
# :. r(r^3-1) = 0#
# :. r=0,1#

From the second equation we get:

# 4theta = pi/2-theta +2npi#
# :. 5theta = pi/2 +4npi/2#
# :. theta = (4n+1)pi/10#

So now we can "assemble" the solutions:

# r= 0=> z=0 #

# r= 1=> z=e^(itheta) #

And By putting #n=0,1,2,3,...# we get:

#theta = (pi)/10, pi/2, (9pi)/10, (13pi)/10, (17pi)/10, ...#

And with these values of #theta#, and using #z=costheta + isintheta # we get:

#n=0: => z = 1/2sqrt(1/2(5+sqrt(5))) + 1/4(sqrt(5)-1)i#
#n=1: => z = 0 + 1i#
#n=2: => z = -1/2sqrt(1/2(5+sqrt(5))) + 1/4(sqrt(5)-1)i#
#n=3: => z = -1/2sqrt(1/2(5-sqrt(5))) - 1/4(sqrt(5)+1)i#
#n=4: => z = 1/2sqrt(1/2(5-sqrt(5))) - 1/4(sqrt(5)+1)i#

After which the pattern repeats. Hence there are six solutions:

# z=0,i #
# z = +-1/2sqrt(1/2(5+sqrt(5))) + 1/4(sqrt(5)-1)i#
# z = +-1/2sqrt(1/2(5-sqrt(5))) - 1/4(sqrt(5)+1)i#
We can see these solutions on the Argand diagram:

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