Complex numbers help

1. A) write down the modules and argument of the complex number -64
I've done that fairly easy (64, pi)
B) hence solve the equation z^4 = -64, giving your answers in the form r (cos x + i sin x) where r > 0 and -pi < x <(or equal to) pi.
I got root 8 (cos pi/4 + i sin pi/4)
C) express each of these four roots in the form a + bi and show, with the aid of a diagram that the points in the complex plane which represent them form the vertices of a square

Really stuck with c

Original post by psycholeo

1. A) write down the modules and argument of the complex number -64
I've done that fairly easy (64, pi)
B) hence solve the equation z^4 = -64, giving your answers in the form r (cos x + i sin x) where r > 0 and -pi < x <(or equal to) pi.
I got root 8 (cos pi/4 + i sin pi/4)
C) express each of these four roots in the form a + bi and show, with the aid of a diagram that the points in the complex plane which represent them form the vertices of a square

Really stuck with c

Where are your other 3 roots from part (B) ??

That's a 4th order polynomial, so it should have 4 roots.

Reply 2

isiaiah d

18

Original post by psycholeo

1. A) write down the modules and argument of the complex number -64
I've done that fairly easy (64, pi)
B) hence solve the equation z^4 = -64, giving your answers in the form r (cos x + i sin x) where r > 0 and -pi < x <(or equal to) pi.
I got root 8 (cos pi/4 + i sin pi/4)
C) express each of these four roots in the form a + bi and show, with the aid of a diagram that the points in the complex plane which represent them form the vertices of a square

Really stuck with c

You should get more answers for B) you have to let
z^4 = r(cos(theta+2kpi)+isin(theta+2kpi)) then use de moivres theorem and vary k for different answers in the range.

For C try to draw in on an argand diagram

Reply 3

psycholeo

OP

9

How can you get more answers for b though:

Z^4 = 64(cos pi + I sin pi)
Z = (64(cos pi + I sin pi))^0.25
= root 8 (cos pi/4 + I sin pi/4)
I can see why it could be - always of that but that's only 2 solutions

Original post by psycholeo

How can you get more answers for b though:

Z^4 = 64(cos pi + I sin pi)
Z = (64(cos pi + I sin pi))^0.25
= root 8 (cos pi/4 + I sin pi/4)
I can see why it could be - always of that but that's only 2 solutions

Because if you have a complex number

z^4

, you can always add on an argument of

2\pi

and land on the exact same complex number again.

Hence

z^4 = 64[\cos(\pi + 2\pi k) + i \sin (\pi + 2\pi k)]

and here

k = -2, -1, 0, 1

for all the roots you need in the required region.

(edited 5 years ago)

Reply 5

psycholeo

OP

9

So z = root 8(cos (pi/4 + kpi/2) + i sin (pi/4 + kpi/2))?

Reply 6

psycholeo

OP

9

And how would you write b?

OP

Original post by psycholeo

And how would you write b?

Not take the 4th root of both equations.

Then you got all your 4 roots.

(edited 5 years ago)

Reply 9

psycholeo

OP

9

But I don't get this. Adding 2 pi gets you into the same complex number. Can you take a picture of how you'd lay it out

Original post by psycholeo

But I don't get this. Adding 2 pi gets you into the same complex number. Can you take a picture of how you'd lay it out

Isn't this technique supposed to be explained in your textbook or something?

z^4 = 64[\cos(\pi + 2\pi k) + i \sin (\pi + 2\pi k)] \implies z = \sqrt{8}[\cos(\frac{\pi}{4} + \frac{k\pi}{2}) + i \sin(\frac{\pi}{4} + \frac{k\pi}{2})]

.

This is true for any

k \in \mathbb{Z}

but it is sufficient to pick any four

k

in a row to obtain all the distinct roots. In order to keep the arguments between

- \pi < \theta < \pi

we need to pick

k = -2, -1, 0, 1

.

So,

Unparseable latex formula:

$\begin{align*} k = -2: \quad & z = \sqrt{8}[\cos(-\frac{3\pi}{4}) + i \sin(-\frac{3\pi}{4})] \\ k = -1: \quad & z = \sqrt{8}[\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4})] \\ k = 0: \quad & z = \ldots \\ k = 1: \quad & z = \ldots \end{align*}$

Those are the four roots.

(edited 5 years ago)

Reply 11

psycholeo

OP

9

Original post by RDKGames

Isn't this technique supposed to be explained in your textbook or something?

z^4 = 64[\cos(\pi + 2\pi k) + i \sin (\pi + 2\pi k)] \implies z = \sqrt{8}[\cos(\frac{\pi}{4} + \frac{k\pi}{2}) + i \sin(\frac{\pi}{4} + \frac{k\pi}{2})]

.

This is true for any

k \in \mathbb{Z}

but it is sufficient to pick any four

k

in a row to obtain all the distinct roots. In order to keep the arguments between

- \pi < \theta < \pi

we need to pick

k = -2, -1, 0, 1

.

So,

Unparseable latex formula:

$\begin{align*} k = -2: \quad & z = \sqrt{8}[\cos(-\frac{3\pi}{4}) + i \sin(-\frac{3\pi}{4})] \\ k = -1: \quad & z = \sqrt{8}[\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4})] \\ k = 0: \quad & z = \ldots \\ k = 1: \quad & z = \ldots \end{align*}$

Those are the four roots.

Why do we choose those particular k values?

Original post by psycholeo

Why do we choose those particular k values?

I explained it.

Original post by RDKGames

In order to keep the arguments between

- \pi < \theta \leq \pi

we need to pick

k = -2, -1, 0, 1

.

The question clearly asks for this region.

Reply 13

Aadilkharl123

17

Im the mathematician around these ends...

<3

Reply 14

psycholeo

OP

9

We don't have a textbook they haven't published it

Reply 15

psycholeo

OP

9

Okay, so the next example is: z^3 = (1 + I) (root 3 - i). Do the normal procedure and you get this in the form z = 6th root of 8 (cos (pi/36 + 2kpi/3)...) in the region -pi < theta < pi

Reply 16

psycholeo

OP

9

Forgot to say that has 4 solutions in that region. So how is that possible?

Reply 17

psycholeo

OP

9

Oh wait no it doesn't nevermind

Original post by psycholeo

Okay, so the next example is: z^3 = (1 + I) (root 3 - i). Do the normal procedure and you get this in the form z = 6th root of 8 (cos (pi/36 + 2kpi/3)...) in the region -pi < theta < pi