Finding square roots of z

Original post by methewthomson

Hi,

I also need help with another question. In the following part, I don't know how to find the two square roots of z:

In part (i) of the question, z was found to be:

I've tried to directly find the square roots, which equaled to 3e(pi/6), but that's not the correct answer. Do you guys know the right method for solving this question?

Thanks

http://madasmaths.com/archive_maths_booklets_further_topics_various.html

file complex_numbers_part_2, look at the start of the sheet

Reply 3

OP

Original post by notnek

Can you post your working? You seem to be missing an

i

unless it's a typo. Also, you should have two square roots.

Thanks you for replying.

Yeah, I guess I missed the "i" when I was posting. Also, I think what I'm asking is how to find the other square root, as you pointed out (I had found this one and misread the mark scheme). So could you please explain how to find the other one?

Thanks

Reply 4

OP

Original post by notnek

I just noticed that r>0 which seems like a mistake in the question if your z is correct.

Are you sure that your answer to part i) is correct? And what answers are given in the mark scheme for ii)?

I'm pasting the required portion from the mark scheme below:

mark scheme.png47.2KB

Reply 5

OP

Original post by TeeEm

http://madasmaths.com/archive_maths_booklets_further_topics_various.html

file complex_numbers_part_2, look at the start of the sheet

I downloaded the file but I'm afraid I can't understand the working for finding the fifth roots (which is similar to, I suppose, finding the square roots.) Can you please explain a bit as to how I should find the other square root in my question? I'd really appreciate your help!

Reply 6

Notnek

21

Original post by methewthomson

I'm pasting the required portion from the mark scheme below:

Sorry I made a mistake in my last post.

Remember that a complex number in the form

re^{i\theta}

can also be written

re^{i(\theta+2k\pi)}

for integers k.

So

z=9e^{(\frac{1}{3}\pi \pm 2\pi) i}

The k values have been chosen as

\pm 1

because of the restricted interval given in the question. Now square rooting:

3e^{(\frac{1}{6}\pi \pm \pi) i}

There is a mistake in the mark-scheme - the line above is correct.

(edited 9 years ago)

Reply 7

Original post by methewthomson

I downloaded the file but I'm afraid I can't understand the working for finding the fifth roots (which is similar to, I suppose, finding the square roots.) Can you please explain a bit as to how I should find the other square root in my question? I'd really appreciate your help!

firstly there are two ways of finding complex square roots

the first method only works for square roots and usually applies to the FP1 papers for most boards

the second method is general and works for all roots and usually applies to the FP2 or FP3 module of most boards

which method are you after?

Reply 8

OP

Original post by TeeEm

firstly there are two ways of finding complex square roots

the first method only works for square roots and usually applies to the FP1 papers for most boards

the second method is general and works for all roots and usually applies to the FP2 or FP3 module of most boards

which method are you after?

Is there a simple method that you could explain? It's actually a question from a level pure methamatics ( not further maths). Thanks.

Reply 9

Original post by methewthomson

Is there a simple method that you could explain? It's actually a question from a level pure methamatics ( not further maths). Thanks.

the method is simple but you need to know the exponential form of complexes..

so z = 9e^(1/3ip) p=pi

complex representation is not unique, you get the same number if you go around the complex plane by 2ip

so z=9e^[1/3ip plus an integer multiple of 2p]
so z=9e^[1/3ip+ i2kp] k = integer

ok so far, shall I continue?

(edited 9 years ago)

Reply 10

sorry about the edits

Reply 11

OP

Original post by TeeEm

the method is simple but you need to know the exponential form of complexes..

so z = 9e^(1/3ip) p=pi

complex representation is not unique, you get the same number if you go around the complex plane by 2ip

so z=9e^[1/3ip plus an integer multiple of 2p]
so z=9e^[1/3ip+ i2kp] k = integer

ok so far, shall I continue?

Yes, I think I understand, so please go on.

Reply 12

Original post by methewthomson

Yes, I think I understand, so please go on.

then you can factorize the exponent a bit

so z=9e^[1/3ip+ i2kp] k = integer

becomes so z=9 e^1/3ip[1+6k] k = integer

then ...

Reply 13

z=9 e^1/3ip[1+6k] k = integer

think of it as indices rules ...

raise both sides to the power of 1/2

z^1/2 = [9 e^1/3ip[1+6k]]^1/2 k = integer

..... then

Reply 14

in the left you have z^1/2

on the right

9^1/2 =3

and

[e^1/3ip[1+6k]]^1/2

=

e^1/6ip[1+6k]

using rule (a^m)^n = a^(mxn) so 1/2*1/3 =1/6

then ....

Reply 15

OP

Original post by TeeEm

then you can factorize the exponent a bit

so z=9e^[1/3ip+ i2kp] k = integer

becomes so z=9 e^1/3ip[1+6k] k = integer

then ...

I'm afraid I don't understand how the "1+6k" part? How does it work?

Reply 16

Original post by methewthomson

I'm afraid I don't understand how the "1+6k" part? How does it work?

write down just this

1/3ip+ i2kp

where p = pi

factorize out

1/3ip

gives

1/3ip(1+6k)

Reply 17

OP

Original post by TeeEm

write down just this

1/3ip+ i2kp

where p = pi

factorize out

1/3ip

gives

1/3ip(1+6k)

I know I'm asking a lot of questions now, but I'm confused about the factorization because I only use factorization in quadratic equations, so I'm not sure how it's done here. I appreciate the time and effort you are putting into this, but could you please explain it a little?

Thanks again

Reply 18