Solve z^5 = 1 + j. Can someone help walk me through this?

z^5 = 1 + j. Just a guide on how to approach and solve this. Thanks. Please walkthrough it in baby steps i am completely new to this.

(edited 7 years ago)

Reply 1

Mr M

Study Forum Helper

Original post by MatthewTG

z^5 = 1 + j. Just a guide on how to approach and solve this. Thanks.

What have you tried?

Reply 2

Naruke

Original post by MatthewTG

z^5 = 1 + j. Just a guide on how to approach and solve this. Thanks.

Express both sides as complex-exponentials

Reply 3

MatthewTG

Original post by Mr M

What have you tried?

I think the de moivre theorum is used but im not sure

Reply 4

the bear

if z = re^iΘ

then z⁵ = r⁵e^i5Θ

and 1 + i = √2e^iπ/4

^sor⁵e^i5Θ= √2e^iπ/4

^{now you should be able to find values for r and}
Θ

Reply 5

Mr M

Study Forum Helper

Original post by MatthewTG

I think the de moivre theorum is used but im not sure

What's the modulus and argument of 1 + j ?

Reply 6

RDKGames

Study Forum Helper

Original post by MatthewTG

I think the de moivre theorum is used but im not sure

De Moivre's theorem states that

(\cos\theta + i\sin \theta)^n = \cos(n\theta) + i \sin(n\theta)

which doesn't really help with it. It might, though it seems a bit long winded.

Anyhow, the best way to approach this is to express your complex number in the form

re^{i\theta}

and then take the 5th root of both sides - keep in mind that you should have 5 solutions as you consider the period of the solutions.

Reply 7

karmacrunch

Original post by Mr M

Good evening, Mr M. Do you know if this is a question I could get in FP1 (OCR MEI)? :smile:

Reply 8

Mr M

Study Forum Helper

Original post by karmacrunch

Good evening, Mr M. Do you know if this is a question I could get in FP1 (OCR MEI)? :smile:

No you couldn't. It's FP2.

Reply 9

karmacrunch

Original post by Mr M

No you couldn't. It's FP2.

Thank you!

Reply 10

atsruser

Original post by the bear

if z = re^iΘ

then z⁵ = r⁵e^i5Θ

and 1 + i = √2e^iπ/4

^sor⁵e^i5Θ= √2e^iπ/4

^{now you should be able to find values for r and}
Θ

You also need to take general form of the argument into account - the argument can be altered by any multiple of

2\pi

without changing the complex number. So we want:

z^5 = \sqrt{2}e^{i(\pi/4+2n\pi)}

for

n \in \mathbb{Z}

else we won't get all of the solutions - there are 5.

Reply 11

the bear

Original post by atsruser

You also need to take general form of the argument into account - the argument can be altered by any multiple of

2\pi

without changing the complex number. So we want:

z^5 = \sqrt{2}e^{i(\pi/4+2n\pi)}

for

n \in \mathbb{Z}

else we won't get all of the solutions - there are 5.

i did not want to make it too easy.

Reply 12

atsruser

Original post by the bear

i did not want to make it too easy.

I see that the bear is a harsh and demanding taskmaster.

Reply 13

the bear

Original post by atsruser

I see that the bear is a harsh and demanding taskmaster.

also bear™ is not a huge fan of j

:mad:

Reply 14

atsruser

Original post by the bear

also bear™ is not a huge fan of j

:mad:

The standard desensitisation therapy for that ailment is to force yourself to read a large number of electronic and electrical engineering texts - in severe cases, the treatment also involves a straitjacket, eyelid separators, and a CD of Beethoven's Ninth symphony.