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

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

Mathematics
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\[\sqrt{\frac{1-i}{2}}\]?
where does the "30" go?
wait,I will use equation

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Other answers:

u=\[\sqrt{\frac{(1-i)}{2}}*30\]
\[[(\sqrt{2}(1-i)/2 ]^{30}\]
aaah much easier!
\[(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)^{30}\]
\[2^\frac{30}{2}*\frac{(1-i)^{30}}{2^{30}}=2^{15}*\frac{(1-i)^{30}}{2^{30}}=\frac{(1-i)^{30}}{2^{15}}\]
makes the idea even easier than what was written
oh dear no my myininaya
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.
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) ?
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
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 :
ok. but if you can write in trig from raising to the power of 30 is easy. maybe this is something else
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
ok this number is \[\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\] \[=\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4})\]
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?
*-(-\[\sqrt{2}/2+\sqrt{2}/2\]
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 \[i\]
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
omg you guys made this totally harder than is should be \[(1-i)*(1-i)=1-2i-i^2=1-2i-(-1)=1-2i+1=-2i\] \[(1-i)^2=-2i\] \[(1-i)^{2*15}=(-2i)^{15}\] \[(1-i)^{30}=(-2i)^{15}\] \[(1-i)^{30}=(-2)^{15}i^{15}=-2^{15}i^{4*3+2}=-2^{15}i^2=-2^{15}(-1)=2^{15}\] \[(1-i)^{30}=2^{15}\] \[(\frac{\sqrt{2}}{2}(1-i))^{30}=\frac{2^{15}}{2^{30}}(1-i)^{30}=\frac{1}{2^{15}}*2^{15}=1\]
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 :)
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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