u=[root2(1-i)/2]30
Find the number u.

- anonymous

u=[root2(1-i)/2]30
Find the number u.

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- anonymous

\[\sqrt{\frac{1-i}{2}}\]?

- anonymous

where does the "30" go?

- anonymous

wait,I will use equation

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## More answers

- myininaya

u=\[\sqrt{\frac{(1-i)}{2}}*30\]

- anonymous

\[[(\sqrt{2}(1-i)/2 ]^{30}\]

- anonymous

aaah much easier!

- anonymous

\[(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)^{30}\]

- myininaya

\[2^\frac{30}{2}*\frac{(1-i)^{30}}{2^{30}}=2^{15}*\frac{(1-i)^{30}}{2^{30}}=\frac{(1-i)^{30}}{2^{15}}\]

- anonymous

makes the idea even easier than what was written

- anonymous

oh dear no my myininaya

- anonymous

i bet it is what i wrote, because that is an easy one. rewrite in trig (polar, whatever) form. then multiply the angle by 30. the modulus (absolute value) is 1 so nothing to worry about since you are on the unit circle.

- myininaya

leave me alone
i don't know math

- myininaya

its all been guesses

- anonymous

wrong answer,I need to found 1-i. :S

- anonymous

ok do you know how to write this in polar form?

- anonymous

because once you do it is really easy

- anonymous

the number is
\[\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\]

- anonymous

which sits right on the unit circle

- anonymous

do you prefer degrees or radians for these?

- anonymous

The answer is 1-i. But I don't know how to find it

- anonymous

actually that is not the answer, but we will find it. it is not hard

- anonymous

but before we begin, do you know how to write this in trigonometric form?

- anonymous

Do you mean A(root2/2,-root2/2) ?

- anonymous

because the idea is just to multiply the angle by 30 and see what you get. the absolute value of this number is 1, and 1 to the power of 30 is 1

- anonymous

no i mean as
\[r(\cos(\theta)+i\sin(\theta))\]

- anonymous

in this case
\[r=1\] because we are right on the unit circle

- anonymous

\[|\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i|=1\]

- anonymous

I haven't learned anything like this yet.

- anonymous

so all we need it
\[\theta\] which should be easy because we see these numbers all the time

- anonymous

well if you have not, you have no way to raise this to the power of 30!

- anonymous

i mean you are not expected to multiply this number by itself 30 times !

- anonymous

Wait,see what is extcally writes :

- anonymous

ok. but if you can write in trig from raising to the power of 30 is easy. maybe this is something else

- anonymous

u=(\[(\sqrt{2/2}\times w)^{30}\]

- anonymous

and we know that w=1-i

- anonymous

yeah well that is what i wrote

- anonymous

The answer is i **

- anonymous

\[(\frac{\sqrt{2}}{2}(1-i))^{30}\] like this yes?

- anonymous

yes

- anonymous

well we can do it easily, but if you have not seen trig from i don't know how to explain it

- anonymous

write it

- anonymous

i mean you certainly are not going to multiply this number by itself 30 times

- anonymous

ok this number is
\[\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\]
\[=\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4})\]

- anonymous

I have an idea. We can do this: (\frac{\sqrt{2}}{2}(1-i))^{29}*\frac{\sqrt{2}}{2}

- anonymous

so when you raise to the power of 30 you just multiply the angle by 30

- anonymous

I have an idea. We can do this: (\frac{\sqrt{2}}{2}(1-i))^{29}*(\frac{\sqrt{2}}{2}(1-i)

- anonymous

omg!

- anonymous

\[(\frac{\sqrt{2}}{2}(1-i))^{29}*\frac{\sqrt{2}}{2}(1-i)\]

- anonymous

ok that is what you wrote yes?

- anonymous

YE!

- anonymous

then what?

- anonymous

*-(-\[\sqrt{2}/2+\sqrt{2}/2\]

- anonymous

ok you have to write
\[\frac{7\pi}{4}\times 30=\frac{105\pi}{2}\] and realize that this is the same place on the unit circle as
\[\frac{\pi}{2}\] and so the answer is
\[i\]

- anonymous

without polar form or trig form i really don't know how you are expected to do this problem.

- anonymous

sorry that was not much help.

- anonymous

thanks

- anonymous

yw

- myininaya

omg you guys made this totally harder than is should be
\[(1-i)*(1-i)=1-2i-i^2=1-2i-(-1)=1-2i+1=-2i\]
\[(1-i)^2=-2i\]
\[(1-i)^{2*15}=(-2i)^{15}\]
\[(1-i)^{30}=(-2i)^{15}\]
\[(1-i)^{30}=(-2)^{15}i^{15}=-2^{15}i^{4*3+2}=-2^{15}i^2=-2^{15}(-1)=2^{15}\]
\[(1-i)^{30}=2^{15}\]
\[(\frac{\sqrt{2}}{2}(1-i))^{30}=\frac{2^{15}}{2^{30}}(1-i)^{30}=\frac{1}{2^{15}}*2^{15}=1\]

- myininaya

i made a typo in first line
\[(1-i)(1-i)=1-2i+i^2=1-2i+(-1)=-2i\]

- myininaya

:)

- anonymous

actually i am not kicking myself yet

- myininaya

lol

- anonymous

since this is wrong

- myininaya

what?

- anonymous

oh wait. i was looking at the first line.

- anonymous

but you are right it must be a "typo"

- anonymous

you do get \[(1-i)(1-i)=-2i\]

- myininaya

yep

- anonymous

the answer is i.

- anonymous

everything looks good but your answer!

- anonymous

because i happen to know that it is i, not 1

- myininaya

i will look at it again

- myininaya

but the approach looks good

- anonymous

taht is ok you just forgot the
\[i^{15}\] in your answer

- anonymous

no it is right, you just forgot one part i think at the end. much much easier in polar form really

- myininaya

\[i^{4*3+3}\]
omg i put a 2 where that 3 was

- anonymous

to hell with IE9!

- myininaya

yes we get i :)

- myininaya

did you see my mistake above
\[(-2)^{15}*i^{15}=-2^{15}*i^{4*3+3}=-2^{15}*i^3=-2^{15}*(-i)=2^{15}*i\]
\[\frac{1}{2^{15}}*(1-i)^{30}=\frac{1}{2^{15}}*(2^{15}*i)=i\]

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