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

where does the "30" go?

wait,I will use equation

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

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

aaah much easier!

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

makes the idea even easier than what was written

oh dear no my myininaya

leave me alone
i don't know math

its all been guesses

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

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

because once you do it is really easy

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

which sits right on the unit circle

do you prefer degrees or radians for these?

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

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

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

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

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

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

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

I haven't learned anything like this yet.

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

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

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

Wait,see what is extcally writes :

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

and we know that w=1-i

yeah well that is what i wrote

The answer is i **

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

yes

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

write it

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

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

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

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

omg!

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

ok that is what you wrote yes?

YE!

then what?

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

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

sorry that was not much help.

thanks

yw

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

:)

actually i am not kicking myself yet

lol

since this is wrong

what?

oh wait. i was looking at the first line.

but you are right it must be a "typo"

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

yep

the answer is i.

everything looks good but your answer!

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

i will look at it again

but the approach looks good

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

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

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

to hell with IE9!

yes we get i :)