anonymous
  • anonymous
u=[root2(1-i)/2]30 Find the number u.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\sqrt{\frac{1-i}{2}}\]?
anonymous
  • anonymous
where does the "30" go?
anonymous
  • anonymous
wait,I will use equation

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