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[SOLVED] How do I solve complex number z^4+16=0

Raerin

Member
Oct 7, 2013
46
How do I solve z^4+16=0 in a+bi form. I don't see how it can be factored so what do I do?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
How do I solve z^4+16=0 in a+bi form. I don't see how it can be factored so what do I do?
Write it as:
$$z^4=-16$$
Write both $z$ and $-16$ in polar form.
Solve, and convert back to $a+bi$ form.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Actually it can be factored since $1 = -i^2 $ we have

\(\displaystyle z^4+16=z^4-i^2 16 = (z^2-4i)(z^2+4i)\)

Now you can consider \(\displaystyle 2(1\pm i)^2= 2(1\pm2i+i^2)=\pm 4i\)
 

Raerin

Member
Oct 7, 2013
46
Now you can consider \(\displaystyle 2(1+i)^2= 2(1+2i+i^2)=4i\)
I don't understand this part.

Is Serena's way easier? Though, I don't really know how to convert it into polar form.
 

Klaas van Aarsen

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Mar 5, 2012
8,780
I don't understand this part.

Is Serena's way easier? Though, I don't really know how to convert it into polar form.
I think my way is easier... :eek:

Let's start with the modulus.
What is the modulus of $-16$?

Now suppose that the modulus of $z$ is $r$.
What should $r$ be?
 

Raerin

Member
Oct 7, 2013
46
I think my way is easier... :eek:

Let's start with the modulus.
What is the modulus of $-16$?

Now suppose that the modulus of $z$ is $r$.
What should $r$ be?
(-16)^2 = sqrt(256) = 16?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I don't understand this part.

Is Serena's way easier? Though, I don't really know how to convert it into polar form.
Apply the expansion for $(x+y)^2=x^2+2xy+y^2$. Since we are looking at the roots of the equation \(\displaystyle z^4=-16\) and since $(z^2-4i)(z^2+4i)=0$ we have either


\(\displaystyle z^2=4i\) or \(\displaystyle z^2=-4i\)

Now since for $z=\sqrt{2}(1+i)$ we have \(\displaystyle z^2=2(1+2i-1)=4i\) then

\(\displaystyle z=\sqrt{2}(1+i)\) is a solution. Similarily we have $z=-\sqrt{2}(1+i)$ as a solution. Using the same manner solve for $z$ the equation \(\displaystyle z^2=-4i\).
 

Klaas van Aarsen

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Mar 5, 2012
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Raerin

Member
Oct 7, 2013
46

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780

Raerin

Member
Oct 7, 2013
46

Klaas van Aarsen

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Mar 5, 2012
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Raerin

Member
Oct 7, 2013
46
Right!

Now suppose $z$ has an angle $\phi$.
Then $z^4$ has an angle $4\phi$.
What should $\phi$ be?
720?

- - - Updated - - -

Apply the expansion for $(x+y)^2=x^2+2xy+y^2$. Since we are looking at the roots of the equation \(\displaystyle z^4=-16\) and since $(z^2-4i)(z^2+4i)=0$ we have either


\(\displaystyle z^2=4i\) or \(\displaystyle z^2=-4i\)

Now since for $z=\sqrt{2}(1+i)$ we have \(\displaystyle z^2=2(1+2i-1)=4i\) then

\(\displaystyle z=\sqrt{2}(1+i)\) is a solution. Similarily we have $z=-\sqrt{2}(1+i)$ as a solution. Using the same manner solve for $z$ the equation \(\displaystyle z^2=-4i\).
Sorry, I still don't fully understand how z=2√(1+i)
 

Klaas van Aarsen

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Mar 5, 2012
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Raerin

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Oct 7, 2013
46

Klaas van Aarsen

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Mar 5, 2012
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oh so the angle is 45?
Yes.

There you go. We now have 1 solution in polar form.
It has modulus $2$ and angle $45^\circ$.
Can you think of a complex number in the form $a+bi$ that corresponds to it?
 

Raerin

Member
Oct 7, 2013
46
Yes.

There you go. We now have 1 solution in polar form.
It has modulus $2$ and angle $45^\circ$.
Can you think of a complex number in the form $a+bi$ that corresponds to it?
z = sqrt(2) + sqrt(2)i?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
To make it a bit more transparent, suppose $w$ is a complex number such that $w^8 = 1$, but no lower power of $w$ is 1.

Then $w^4$ is a square root of 1, and since $w^4 \neq 1$, $w^4 = -1$.

Hence:

$z^4 + 16 = z^4 - 16w^4 = (z^2 + 4w^2)(z^2 - 4w^2)$

$= (z^2 - 4(iw)^2)(z + 2w)(z - 2w) = (z + 2iw)(z - 2iw)(z + 2w)(z - 2w)$.

So now we need to find $w$.

As indicated above, $w$ is a root of $x^8 - 1$. Let's factor $x^8 - 1$.

$x^8 - 1 = (x^4 + 1)(x^4 - 1)$.

Now $x^4 - 1 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x - 1)(x + 1)$.

The roots -1, and 1 aren't what we're looking for, and the roots of $x^2 + 1$ are $i$ and $-i$, so $w$ must be a root of $x^4 + 1$ (because the fourth roots of 1, that is the roots of $x^4 - 1$ all have some lower power than 8 equal to 1).

Writing:

$x^4 + 1 = x^4 - i^2 = (x^2 + i)(x^2 - i)$

It's clear that $w^2 = \pm i$.

So what we are really looking for is a square root of $i$ or $-i$.

So suppose $(a + bi)^2 = i$.

Expanding, we have:

$a^2 - b^2 + 2abi = i$, so:

$a^2 - b^2 = 0$
$2ab = 1$

From the second equation, we have:

$b = \dfrac{1}{2a}$. Clearly $a \neq 0$ (or else $2ab = 0$).

Substituting this in the first equation we get:

$a^2 - \dfrac{1}{4a^2} = 0$.

Multiplying by $a^2$ we get:

$a^4 = \dfrac{1}{4}$.

We don't need all possible solutions to this, just one (out of the 4 possible), so take square roots of both sides:

$a^2 = \dfrac{1}{2}$

and again:

$a = \dfrac{\sqrt{2}}{2}$.

Then $b = \dfrac{1}{2a} = \dfrac{1}{2}\cdot\dfrac{1}{a} = \dfrac{\sqrt{2}}{2}$ as well.

Hence one possible solution is:

$w = \dfrac{\sqrt{2} + \sqrt{2}i}{2}$,

and the roots of $z^4 + 16$ can now be calculated:

$r_1 = 2w = \sqrt{2} + \sqrt{2}i$
$r_2 = -2w = -\sqrt{2} - \sqrt{2}i$
$r_3 = 2iw = -\sqrt{2} + \sqrt{2}i$
$r_4 = -2iw = \sqrt{2} - \sqrt{2}i$

If you feel like doing extra work, show that any of the 3 other choices for $a$ when we solved:

$a^4 = \dfrac{1}{4}$ lead to the same 4 solutions to $z^4 + 16 = 0$.

******

In light of ILikeSerena's posts, note that writing:

$e^{i\theta} = \cos\theta + i\sin\theta$

makes it clear that one way to easily get $w$ is to choose:

$w = e^{i2\pi/8} = e^{i\pi/4} = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})$

(since $w^8 = (e^{i2\pi/8})^8 = e^{i2\pi} = \cos(2\pi) + i \sin(2\pi) = 1 + i0 = 1$).

which of course leads to the same answer.

His approach also indicates a strategy for finding $n$-th roots of a complex number $z$ in general:

1) Divide by the magnitude to "pull back" to the unit circle.
2) Use trigonometry to convert to polar form (typically this involves some use of the arctan function, but some angles are obvious).
3) Divide the angle of $z$ by $n$, to get the angle of $\sqrt[n]{z}$ (which has the same angle as $z/|z|$ which lies on the unit circle).
4) Convert to rectangular coordinates ($x + yi$) using trigonometry.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
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z = sqrt(2) + sqrt(2)i?
Yes! :D
Do you recognize the value for $z$ that ZaidAlyafey came up with?
Perhaps you might calculate $(\sqrt 2 + (\sqrt 2)i)^4$ to verify that it really is a solution?


Btw, there are 3 more solutions.
That is because $4\phi=180^\circ$ has more solutions.
If $4\phi = 360^\circ + 180^\circ$, we have the same angle but a different $\phi$...
 

Raerin

Member
Oct 7, 2013
46
Yes! :D
Do you recognize the value for $z$ that ZaidAlyafey came up with?
Perhaps you might calculate $(\sqrt 2 + (\sqrt 2)i)^4$ to verify that it really is a solution?


Btw, there are 3 more solutions.
That is because $4\phi=180^\circ$ has more solutions.
If $4\phi = 360^\circ + 180^\circ$, we have the same angle but a different $\phi$...
Yep, I did! I just didn't understand how they got to that conclusion from z^2=4i
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,780
Yep, I did! I just didn't understand how they got to that conclusion from z^2=4i
Ah, I suspect he secretly converted to polar coordinates and back, and then presented the result.
At least that is how I would do it. :rolleyes:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Another approach would be to use de Moivre's theorem.

\(\displaystyle z^4=-16=2^4\left(\cos\left((2k+1)\pi \right)+i\sin\left((2k+1)\pi \right) \right)\) where \(\displaystyle k\in\mathbb{Z}\)

Hence:

\(\displaystyle z=2\left(\cos\left((2k-1)\frac{\pi}{4} \right)+i\sin\left((2k-1)\frac{\pi}{4} \right) \right)\) where \(\displaystyle k\in\{0,1,2,3\}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Squaring \(\displaystyle \sqrt{\frac{x}{2}}(1+i)\), we can generate any purely imaginary complex number.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Squaring \(\displaystyle \sqrt{\frac{x}{2}}(1+i)\), we can generate any purely imaginary complex number.
Sort of...

Say we have $bi$, where $b > 0$ is real.

Suppose we have "some" square root of $i$ (there are two, and it does not matter which one we choose, as there is no sense of "positive" in the complex numbers). call this square root of $i, z_0 = x + yi$.

Then $(\sqrt{b}z_0)^2 = (\sqrt{b})^2(z_0)^2 = bi$.

It turns out that choosing:

$x = y = \dfrac{\sqrt{2}}{2}$ works, but so does choosing:

$x = y = -\dfrac{\sqrt{2}}{2}$

(in ILike Serena's view, this is because:

$2\cdot \dfrac{\pi}{4} = \dfrac{\pi}{2}$ as well as:

$2 \cdot \dfrac{5\pi}{4} = \dfrac{5\pi}{2} = \dfrac{\pi}{2}\text{ (mod }2\pi)$).

If, however, $b < 0$, it makes more sense to use a square root of $-i$, that is, to use:

$\sqrt{|b|}\dfrac{\sqrt{2}}{2}(-1 + i)$, or:

$\sqrt{|b|}\dfrac{\sqrt{2}}{2}(1 - i)$.

Note that while your assertion is still correct even when $x < 0$, the term $\sqrt{x}$ is ambiguous in such a case, as both $i$ and $-i$ are square roots of -1, and there is no way to tell them apart algebraically or geometrically ($\Bbb R^2$ does not come equipped with a "preferred orientation" and the map $a + ib \mapsto a - ib$ is a field automorphism).

I want to stress this, because it is important:

$\dfrac{1}{\sqrt{2}}(1 + i)$ is not THE square root of $i$, it is A square root of $i$.

Defining a single-valued function:

$z \mapsto z^{1/2}$ is problematic in the complex field for this very reason (similar to the problems in defining a logarithm function, which is actually directly related to this same issue).

We can solve this problem by choosing a *principal branch*, but this is a CHOICE, not something imposed on us (like by an order, which is the case in the real numbers) by the internal structure of the complex field. We can also solve this by using a Riemann surface as the image space, neatly avoiding creating a "multi-valued function" (something of an oxymoron).

For similar reasons, it is dangerous to think of $i$ as "the" square root of -1, it is just "one" of the two square roots. Euler's formula, and the related theorem of DeMoivre skirts this issue by using an (arbitrarily chosen) orientation induced by choosing "counterclockwise" as the direction of "increasing angle"....one can easily imagine a parallel universe to ours where mathematicians decided on "clockwise" instead, and there is no reason save for historical convention to use this orientation (just as there is no intrinsic reason for making the $x$-axis "horizontal", or to make leftwards the increasing direction on the $x$-axis).

In my previous post, I tried to make it clear that from an algebraic standpoint, all 4 solutions to:

$z^4 + 16 = 0$

are equally valid, there is no "preferred root".