Alexis1994_13
u=[root2(1-i)/2]30
Find the number u.
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anonymous
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\[\sqrt{\frac{1-i}{2}}\]?
anonymous
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where does the "30" go?
Alexis1994_13
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wait,I will use equation
myininaya
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u=\[\sqrt{\frac{(1-i)}{2}}*30\]
Alexis1994_13
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\[[(\sqrt{2}(1-i)/2 ]^{30}\]
anonymous
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aaah much easier!
anonymous
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\[(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)^{30}\]
myininaya
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\[2^\frac{30}{2}*\frac{(1-i)^{30}}{2^{30}}=2^{15}*\frac{(1-i)^{30}}{2^{30}}=\frac{(1-i)^{30}}{2^{15}}\]
anonymous
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makes the idea even easier than what was written
anonymous
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oh dear no my myininaya
anonymous
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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
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leave me alone
i don't know math
myininaya
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its all been guesses
Alexis1994_13
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wrong answer,I need to found 1-i. :S
anonymous
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ok do you know how to write this in polar form?
anonymous
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because once you do it is really easy
anonymous
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the number is
\[\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\]
anonymous
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which sits right on the unit circle
anonymous
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do you prefer degrees or radians for these?
Alexis1994_13
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The answer is 1-i. But I don't know how to find it
anonymous
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actually that is not the answer, but we will find it. it is not hard
anonymous
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but before we begin, do you know how to write this in trigonometric form?
Alexis1994_13
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Do you mean A(root2/2,-root2/2) ?
anonymous
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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
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no i mean as
\[r(\cos(\theta)+i\sin(\theta))\]
anonymous
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in this case
\[r=1\] because we are right on the unit circle
anonymous
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\[|\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i|=1\]
Alexis1994_13
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I haven't learned anything like this yet.
anonymous
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so all we need it
\[\theta\] which should be easy because we see these numbers all the time
anonymous
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well if you have not, you have no way to raise this to the power of 30!
anonymous
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i mean you are not expected to multiply this number by itself 30 times !
Alexis1994_13
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Wait,see what is extcally writes :
anonymous
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ok. but if you can write in trig from raising to the power of 30 is easy. maybe this is something else
Alexis1994_13
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u=(\[(\sqrt{2/2}\times w)^{30}\]
Alexis1994_13
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and we know that w=1-i
anonymous
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yeah well that is what i wrote
Alexis1994_13
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The answer is i **
anonymous
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\[(\frac{\sqrt{2}}{2}(1-i))^{30}\] like this yes?
Alexis1994_13
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yes
anonymous
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well we can do it easily, but if you have not seen trig from i don't know how to explain it
Alexis1994_13
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write it
anonymous
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i mean you certainly are not going to multiply this number by itself 30 times
anonymous
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ok this number is
\[\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\]
\[=\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4})\]
Alexis1994_13
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I have an idea. We can do this: (\frac{\sqrt{2}}{2}(1-i))^{29}*\frac{\sqrt{2}}{2}
anonymous
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so when you raise to the power of 30 you just multiply the angle by 30
Alexis1994_13
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I have an idea. We can do this: (\frac{\sqrt{2}}{2}(1-i))^{29}*(\frac{\sqrt{2}}{2}(1-i)
Alexis1994_13
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omg!
anonymous
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\[(\frac{\sqrt{2}}{2}(1-i))^{29}*\frac{\sqrt{2}}{2}(1-i)\]
anonymous
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ok that is what you wrote yes?
Alexis1994_13
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YE!
anonymous
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then what?
Alexis1994_13
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*-(-\[\sqrt{2}/2+\sqrt{2}/2\]
anonymous
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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
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without polar form or trig form i really don't know how you are expected to do this problem.
anonymous
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sorry that was not much help.
Alexis1994_13
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thanks
anonymous
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yw
myininaya
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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
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i made a typo in first line
\[(1-i)(1-i)=1-2i+i^2=1-2i+(-1)=-2i\]
myininaya
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:)
anonymous
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actually i am not kicking myself yet
myininaya
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lol
anonymous
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since this is wrong
myininaya
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what?
anonymous
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oh wait. i was looking at the first line.
anonymous
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but you are right it must be a "typo"
anonymous
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you do get \[(1-i)(1-i)=-2i\]
myininaya
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yep
Alexis1994_13
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the answer is i.
anonymous
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everything looks good but your answer!
anonymous
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because i happen to know that it is i, not 1
myininaya
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i will look at it again
myininaya
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but the approach looks good
anonymous
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taht is ok you just forgot the
\[i^{15}\] in your answer
anonymous
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no it is right, you just forgot one part i think at the end. much much easier in polar form really
myininaya
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\[i^{4*3+3}\]
omg i put a 2 where that 3 was
anonymous
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to hell with IE9!
myininaya
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yes we get i :)
myininaya
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did you see my mistake above
\[(-2)^{15}*i^{15}=-2^{15}*i^{4*3+3}=-2^{15}*i^3=-2^{15}*(-i)=2^{15}*i\]
\[\frac{1}{2^{15}}*(1-i)^{30}=\frac{1}{2^{15}}*(2^{15}*i)=i\]