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[SOLVED] Complex Number equation

Raerin

Member
Oct 7, 2013
46
Solve the following equations in the form a +bi.
a) z^3-1=0
b) 3z^4+i=1-2i

Apparently, the solution for a) is this:
z^3=1
z=1
z=a+bi=1
sqrt(a^2+b^2)=1

I don't understand why a+bi=1 is not the final answer. Why do you have to make it into a modulus?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Solve the following equations in the form a +bi.
a) z^3-1=0
b) 3z^4+i=1-2i

Apparently, the solution for a) is this:
z^3=1
z=1
z=a+bi=1
sqrt(a^2+b^2)=1

I don't understand why a+bi=1 is not the final answer. Why do you have to make it into a modulus?
To solve a polynomial equation in complex numbers, there are always as many roots as the order of the polynomial. I'd advise simplifying as much as possible (e.g. up to z^3 = something), writing the RHS in a general exponential form, and going from there...
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Solve the following equations in the form a +bi.
a) z^3-1=0
b) 3z^4+i=1-2i

Apparently, the solution for a) is this:
z^3=1
z=1
z=a+bi=1
sqrt(a^2+b^2)=1

I don't understand why a+bi=1 is not the final answer. Why do you have to make it into a modulus?
There are 3 solutions for a).

Every complex number can be written as the combination of the modulus and the angle (aka its "argument").
If we multiply 2 complex numbers, the result has a modulus that is the product of the 2 moduli, and it has an angle that is the sum of the 2 angles.

Suppose the modulus of $z$ is $r$, and the angle of $z$ is $\phi$.
Then the modulus of $z^3$ is $r^3$, and its angle is $3\phi$.

To solve $z^3=1$, we need that $r^3 = 1$ and that $3\phi = 2n\pi$ (where $n$ is a whole number).
It follows that the modulus $r$ must be $1$.
And that the angle $\phi = 0,\ 2\pi/3,\ 4\pi/3$.

In other words, the solutions are:
\begin{aligned}
z&=1 \\
z&=-\frac 1 2 + \frac 1 2 \sqrt 3 i \\
z&=-\frac 1 2 - \frac 1 2 \sqrt 3 i
\end{aligned}
 

Raerin

Member
Oct 7, 2013
46
There are 3 solutions for a).

Every complex number can be written as the combination of the modulus and the angle (aka its "argument").
If we multiply 2 complex numbers, the result has a modulus that is the product of the 2 moduli, and it has an angle that is the sum of the 2 angles.

Suppose the modulus of $z$ is $r$, and the angle of $z$ is $\phi$.
Then the modulus of $z^3$ is $r^3$, and its angle is $3\phi$.

To solve $z^3=1$, we need that $r^3 = 1$ and that $3\phi = 2n\pi$ (where $n$ is a whole number).
It follows that the modulus $r$ must be $1$.
And that the angle $\phi = 0,\ 2\pi/3,\ 4\pi/3$.

In other words, the solutions are:
\begin{aligned}
z&=1 \\
z&=-\frac 1 2 + \frac 1 2 \sqrt 3 i \\
z&=-\frac 1 2 - \frac 1 2 \sqrt 3 i
\end{aligned}
So how am I supposed to write that in a+bi form?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
So how am I supposed to write that in a+bi form?
The first solution has $a=1,\ b=0$.
The second solution has $a=-\frac 1 2,\ b=\frac 1 2 \sqrt 3$.

In other words, the solutions I gave are in $a+bi$ form.
 

Raerin

Member
Oct 7, 2013
46
The first solution has $a=1,\ b=0$.
The second solution has $a=-\frac 1 2,\ b=\frac 1 2 \sqrt 3$.

In other words, the solutions I gave are in $a+bi$ form.
Oh, I see, I was under the impression that I only need one answer in a+bi form.

Thanks!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Oh, I see, I was under the impression that I only need one answer in a+bi form.

Thanks!
Every complex number has 2 representations: the $a+bi$ or cartesian form and the $r^{}e^{i\phi}$ or polar form.
In this case they are asking for the cartesian form.