# Thread: All solutions - Complex Number

1. ## All solutions - Complex Number

$z^4+16=0$. Solve and graph all solutions.
Hint: Then use the solutions to factor $z^4+16=0$ into quadratic factors with real coefficients.

It gave me a hint, but I don't understand this hint.

For me, now I know only $z = -2i$. And after that, I am not sure how to make it in $z=re^{i\theta}$

Would you please tell me the direction.
Thank you.

2. Look here.
In my last reply, there is a general way of finding roots.

3. Originally Posted by noppawit
$z^4+16=0$. Solve and graph all solutions.
Hint: Then use the solutions to factor $z^4+16=0$ into quadratic factors with real coefficients.

It gave me a hint, but I don't understand this hint.

For me, now I know only $z = -2i$. And after that, I am not sure how to make it in $z=re^{i\theta}$

Would you please tell me the direction.
Thank you.
Hello : Pose : Z2 = t

4. Originally Posted by noppawit
$z^4+16=0$. Solve and graph all solutions.
Hint: Then use the solutions to factor $z^4+16=0$ into quadratic factors with real coefficients.

It gave me a hint, but I don't understand this hint.

For me, now I know only $z = -2i$. And after that, I am not sure how to make it in $z=re^{i\theta}$

Would you please tell me the direction.
Thank you.
$z = -2i$ isn't one of the solutions. $z^4 = (-2i)^4 = 16$.

This is what I get:

\begin{aligned}
z^4 + 16 &= 0 \\
z^4 &= -16
\end{aligned}

$-16 = 16e^{i\pi}$

If $z = re^{i\theta}$, then the nth roots of a complex number would be in the form of

$\sqrt[n]{z} = \sqrt[n]{r}\cdot e^{\frac{\theta + 2\pi k}{n}i}$ ,
where k = 0, 1, 2, ... n - 1.

In our case, n = 4, r = 16, and k will go from 0 to 3. The angle simplifies to
$\frac{\theta + 2\pi k}{n} = \frac{\pi + 2\pi k}{4} = \frac{\pi}{4} + \frac{\pi k}{2}$

The four fourth roots of $-16 = 16e^{i\pi}$ are:

\begin{aligned}
z_1 &= \sqrt[4]{16}\cdot e^{\left(\frac{\pi}{4} + \frac{0\pi}{2}\right)i} \\
&= 2e^{\frac{\pi}{4}i} \\
&= 2\left[\cos\left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right] \\
&= 2\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right) \\
&= \sqrt{2} + \sqrt{2}i
\end{aligned}

\begin{aligned}
z_2 &= \sqrt[4]{16}\cdot e^{\left(\frac{\pi}{4} + \frac{\pi}{2}\right)i} \\
&= 2e^{\frac{3\pi}{4}i} \\
&= 2\left[\cos\left(\frac{3\pi}{4}\right) + i \sin \left(\frac{3\pi}{4}\right)\right] \\
&= 2\left(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right) \\
&= -\sqrt{2} + \sqrt{2}i
\end{aligned}

\begin{aligned}
z_3 &= \sqrt[4]{16}\cdot e^{\left(\frac{\pi}{4} + \frac{2\pi}{2}\right)i} \\
&= 2e^{\frac{5\pi}{4}i} \\
&= 2\left[\cos\left(\frac{5\pi}{4}\right) + i \sin \left(\frac{5\pi}{4}\right)\right] \\
&= 2\left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right) \\
&= -\sqrt{2} - \sqrt{2}i
\end{aligned}

\begin{aligned}
z_4 &= \sqrt[4]{16}\cdot e^{\left(\frac{\pi}{4} + \frac{3\pi}{2}\right)i} \\
&= 2e^{\frac{7\pi}{4}i} \\
&= 2\left[\cos\left(\frac{7\pi}{4}\right) + i \sin \left(\frac{7\pi}{4}\right)\right] \\
&= 2\left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right) \\
&= \sqrt{2} - \sqrt{2}i
\end{aligned}

Furthermore, I don't think $z^4+16=0$ can be factored into quadratic factors with real coefficients. $z^4+16 = (z^2 + 4i)(z^2 - 4i)$, but that doesn't work.

Are you sure the problem isn't $z^4 {\color{red}-} 16=0$ (with a minus sign)?

01

5. Thank you for your reply. Anyway, $z^4+16=0$ is a correct one. This problem came from "Advance Engineering Mathematics 9th edition", from Wiley publisher.

6. Originally Posted by yeongil

Furthermore, I don't think $z^4+16=0$ can be factored into quadratic factors with real coefficients. $z^4+16 = (z^2 + 4i)(z^2 - 4i)$, but that doesn't work.
Your intuition was right; you can factor it like you did, and it actually makes things quite easy. Your four roots become $\pm\sqrt{4i}, \pm i\sqrt{4i}$. Notice that you have

$\sqrt{i} = \exp(\pi i/4) = \frac{1+i}{\sqrt 2}$

7. Originally Posted by yeongil
[snip]
Furthermore, I don't think $z^4+16=0$ can be factored into quadratic factors with real coefficients. $z^4+16 = (z^2 + 4i)(z^2 - 4i)$, but that doesn't work.

[snip]
Not so:

$z^4 + 16 = z^4 + 8z^2 - 8z^2 + 16 = (z^4 + 8z^2 + 16) - 8z^2 = (z^2 + 4)^2 - 8z^2$

$= (z^2 + 4 - 2 \sqrt{2} z) (z^2 + 4 + 2 \sqrt{2} z)$

$= (z^2 - 2 \sqrt{2} z + 4) (z^2 + 2 \sqrt{2} z + 4)$.

8. ^ Quite clever! Would have never thought of that!

01

9. Originally Posted by yeongil
In our case, n = 4, r = 16, and k will go from 0 to 3. The angle simplifies to
$\frac{\theta + 2\pi k}{n} = \frac{\pi + 2\pi k}{4} = \frac{\pi}{4} + \frac{\pi k}{2}$
From this, why $\theta = \pi$?

10. Originally Posted by noppawit
From this, why $\theta = \pi$?
Hi noppawit,

Note that we have $z^4=-16$ therefore we are trying to find the 4th roots of $-16$ and so $\theta = \arg(-16) = \pi$

Hope this helps.