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#### Raerin

##### Member

- Oct 7, 2013

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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?

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- #1

- Oct 7, 2013

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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?

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- #2

- Mar 5, 2012

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Write it as: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?

$$z^4=-16$$

Write both $z$ and $-16$ in polar form.

Solve, and convert back to $a+bi$ form.

- Jan 17, 2013

- 1,667

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

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

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- #4

- Oct 7, 2013

- 46

I don't understand this part.Now you can consider \(\displaystyle 2(1+i)^2= 2(1+2i+i^2)=4i\)

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

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- #5

- Mar 5, 2012

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I think my way is easier...I don't understand this part.

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

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?

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- #6

- Oct 7, 2013

- 46

(-16)^2 = sqrt(256) = 16?I think my way is easier...

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?

- Jan 17, 2013

- 1,667

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 eitherI don't understand this part.

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

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

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- #8

- Mar 5, 2012

- 8,780

Yep.(-16)^2 = sqrt(256) = 16?

Now we're looking for a modulus $r$ for $z$ such that $r^4 = 16$...

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- #9

- Oct 7, 2013

- 46

r=2Yep.

Now we're looking for a modulus $r$ for $z$ such that $r^4 = 16$...

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- #10

- Mar 5, 2012

- 8,780

Exactly!

Now let's look at the angle.

Which angle would the number $-16$ have?

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- #11

- Oct 7, 2013

- 46

180 degrees?Exactly!

Now let's look at the angle.

Which angle would the number $-16$ have?

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- #12

- Mar 5, 2012

- 8,780

Right!180 degrees?

Now suppose $z$ has an angle $\phi$.

Then $z^4$ has an angle $4\phi$.

What should $\phi$ be?

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- #13

- Oct 7, 2013

- 46

720?Right!

Now suppose $z$ has an angle $\phi$.

Then $z^4$ has an angle $4\phi$.

What should $\phi$ be?

- - - Updated - - -

Sorry, I still don't fully understand how z=2√(1+i)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 \(\displaystyle z^2=-4i\).

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- #14

- Mar 5, 2012

- 8,780

Not quite.720?

We have that $4\phi=180^\circ$.

Does that work with $\phi=720^\circ$?

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- #15

- Oct 7, 2013

- 46

oh so the angle is 45?Not quite.

We have that $4\phi=180^\circ$.

Does that work with $\phi=720^\circ$?

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- #16

- Mar 5, 2012

- 8,780

Yes.oh so the angle is 45?

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?

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- #17

- Oct 7, 2013

- 46

z = sqrt(2) + sqrt(2)i?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?

- Feb 15, 2012

- 1,967

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.

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- #19

- Mar 5, 2012

- 8,780

Yes!z = sqrt(2) + sqrt(2)i?

Do you recognize the value for $z$ that

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 $\phi$...

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- #20

- Oct 7, 2013

- 46

Yep, I did! I just didn't understand how they got to that conclusion from z^2=4iYes!

Do you recognize the value for $z$ thatZaidAlyafeycame 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 $\phi$...

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- #21

- Mar 5, 2012

- 8,780

Ah, I suspect he secretly converted to polar coordinates and back, and then presented the result.Yep, I did! I just didn't understand how they got to that conclusion from z^2=4i

At least that is how I would do it.

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- #22

\(\displaystyle z^4=-16=2^4\left(\cos\left((2k+1)\pi \right)+i\sin\left((2k+1)\pi \right) \right)\) 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 \(\displaystyle k\in\{0,1,2,3\}\)

- Jan 17, 2013

- 1,667

- Feb 15, 2012

- 1,967

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".