Question

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

Answer 1

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

Answer 2

where does the "30" go?

Answer 3

wait,I will use equation

Answer 4

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

Answer 5

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

Answer 6

aaah much easier!

Answer 7

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

Answer 8

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

Answer 9

makes the idea even easier than what was written

Answer 10

oh dear no my myininaya

Answer 11

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.

Answer 12

leave me alone
i don't know math

Answer 13

its all been guesses

Answer 14

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

Answer 15

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

Answer 16

because once you do it is really easy

Answer 17

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

Answer 18

which sits right on the unit circle

Answer 19

do you prefer degrees or radians for these?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

I haven't learned anything like this yet.

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

Wait,see what is extcally writes :

Answer 33

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

Answer 34

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

Answer 35

and we know that w=1-i

Answer 36

yeah well that is what i wrote

Answer 37

The answer is i **

Answer 38

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

Answer 39

yes

Answer 40

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

Answer 41

write it

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

omg!

Answer 48

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

Answer 49

ok that is what you wrote yes?

Answer 50

YE!

Answer 51

then what?

Answer 52

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

Answer 53

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
\[i\]

Answer 54

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

Answer 55

sorry that was not much help.

Answer 56

thanks

Answer 57

yw

Answer 58

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

Answer 59

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

Answer 60

:)

Answer 61

actually i am not kicking myself yet

Answer 62

lol

Answer 63

since this is wrong

Answer 64

what?

Answer 65

oh wait. i was looking at the first line.

Answer 66

but you are right it must be a "typo"

Answer 67

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

Answer 68

yep

Answer 69

the answer is i.

Answer 70

everything looks good but your answer!

Answer 71

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

Answer 72

i will look at it again

Answer 73

but the approach looks good

Answer 74

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

Answer 75

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

Answer 76

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

Answer 77

to hell with IE9!

Answer 78

yes we get i :)

Answer 79

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