need some help with complex numbers :( ! (1 Viewer)

Validity · Jan 28, 2012

Could anyone help me with the following questions please

(Terry Lee 2.1 Q5)
Find the three cube roots of : (anyone know how to do this without demoivres theorem because the theorem isn't covered before this)
a) 1 b) -1

(Terry Lee 2.2 Q11)
Find square roots of

i don't get what to do

)
a) 5+2i

(Terry Lee 2.2 Q13c)
Express the following in the form a+ib where a and b are real:

Random Question:
i don't get the arg thing when doing the mod-arg form, how do we know what value of 'k' to multiply when using:

THANK YOU EVERYONE THAT IS WILLING TO HELP! =)

IamBread · Jan 28, 2012

2.1 Q5

$$For the first one, you have $ w^3 = 1$

$w^3 - 1=0$

$(w-1)(w^2 + w +1)=0$

$Solve the quadratic to give you the roots, same with w^3 = -1$

2.2 Q11

$$Next one, Let $ a +bi = \sqrt{5 + 2i}$

$a^2 - b^2 + 2abi = 5 + 2i$

$$Equating real and imaginary parts, $ 2ab = 2$

$a = \frac{1}{b}, b = \frac{1}{a}$

$a^2 - b^2 = 5$

$$From the line above.. $ a^2 - (\frac{1}{a})^2 - 5 = 0$

$a^4 - 5a^2 - 1 = 0$

$a^2 = \frac{5 \pm \sqrt{25 + 4}}{2}$

$a^2 = \frac{5 \pm \sqrt{29}}{2}$

$$We take the positive answer for $ a^2$ as $ a $ is real, $ a = \pm\sqrt{\frac{5 + \sqrt{29}}{2}}$

$$As $b = \frac{1}{a}, b = \pm\frac{1}{\sqrt{\frac{5 + \sqrt{29}}{2}}}$

$\therefore \sqrt{5 + 2i} = \pm(\sqrt{\frac{5 + \sqrt{29}}{2}} + \frac{1}{\sqrt{\frac{5 + \sqrt{29}}{2}}}i)$

$$Alternatively you can use DeMoivres theorem, which probably would have been easier now that I think about it lol.$

And AAEldar covered the rest

AAEldar · Jan 28, 2012

2.1 Q5

$z^3=1 \rightarrow z^3-1=0$ and factorise. Likewise for b)

2.2 Q11

Use De Moivre's. The answer won't be pretty, but you have to make $z^2=5+2i$ and convert to polar co-ordinates. Or what IamBread did... haha.

2.2 Q13c

$\frac{1}{1+\cos\theta+i\sin\theta}\times\frac{1+\cos\theta-i\sin\theta}{1+\cos\theta-i\sin\theta}$

$\frac{1}{1-\cos\theta-i\sin\theta}\times\frac{1-\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}$

Those should work.

Random Question

k should be an integer, i.e. 0, 1, 2, 3...

Validity · Jan 28, 2012

IamBread said:
$$We take the positive answer for $ a^2$ as a is real, $ a = \pm\sqrt{\frac{5 + \sqrt{29}}{2}}$

umm can you explain how taking the negative answer would make it unreal? is it ALWAYS the positive to make it real?

AAEldar said:
k should be an integer, i.e. 0, 1, 2, 3...

how do we know what integer though? like for 150 degrees they -1 pi and sometimes they would + 2pi for other things .

Carrotsticks · Jan 28, 2012

Validity said:
umm can you explain how taking the negative answer would make it unreal? is it ALWAYS the positive to make it real?

EDIT: Did not see the lack of plus/minus.

Validity said:
how do we know what integer though? like for 150 degrees they -1 pi and sometimes they would + 2pi for other things .

It depends on the question.

pokka · Jan 28, 2012

Validity said:
umm can you explain how taking the negative answer would make it unreal? is it ALWAYS the positive to make it real?

how do we know what integer though? like for 150 degrees they -1 pi and sometimes they would + 2pi for other things .

So to answer your first question, if you took the negative, a is unreal. Remember if "a" is real then a^2 is positive and real. Type $5-\sqrt{29}$ into you calculator and you will find that the answer is negative so you HAVE to accept the positive answer for a^2.
I can't seem to see what you "random question" is so i can't answer you just yet but i'll get back to you on that...(or someone else will)

IamBread · Jan 29, 2012

Validity said:
umm can you explain how taking the negative answer would make it unreal? is it ALWAYS the positive to make it real?.

a is a real number which means a^2 must be positive as if a^2 is negative then a would be nonreal

IamBread · Jan 29, 2012

Validity said:
how do we know what integer though? like for 150 degrees they -1 pi and sometimes they would + 2pi for other things .

$$It's for all integers, unless bounded. Say for example we have a question, what is $ sin^{-1}0$

$$Well there are an infinite number of solutions, so we write it as $ \pm k \pi, k \in \mathbb{Z}.$

$That means the solutions are $ 0, \pm\pi, \pm2\pi, \pm3\pi, \pm4\pi $ etc etc$

AAEldar · Jan 29, 2012

$\mathbb{Z}$

\mathbb{Z}

Carrotsticks · Jan 29, 2012

IamBread said:
$That means the solutions are $ 0, \pm \pi, \pm 2\pi, \pm 3\pi, \pm 4\pi $ etc etc$

Plus/minus.

IamBread · Jan 29, 2012

AAEldar said:
$\mathbb{Z}$

\mathbb{Z}

Thanks

Carrotsticks said:
Plus/minus.

Lol I forgot to do them, I did have $\pm k \pi$ though

Peeik · Jan 29, 2012

should i be surprised spiral hasnt replied?

SpiralFlex · Jan 29, 2012

Peeik said:
should i be surprised spiral hasnt replied?

Why? It's already answered. I AM WATCHING TENNIS.

need some help with complex numbers :( ! (1 Viewer)

Validity

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AAEldar

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pokka

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