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

##### 2 Answers

#### Answer:

#### Explanation:

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

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

Remembering that

Now to solve for the radius, we know that:

By taking the absolute value of both sides we have

Which we can split

But

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

#### 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

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

# {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

# {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

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

And with these values of

#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: