Expert: Paul Klarreich - 2/24/2008

Question
Hello!

Ok, so this is what I have to do. I have to make a chart of the real and imaginary numbers and graph them on the basis of the results I obtain from these problems. I have to calculate 10^[i{(1/2)^10}], 10^[i{(1/2)^9}],10^[i{(1/2)^8}] and so forth until 1/2^0. Then I must calculate 10^[i/8], then 10^[i/4], and use those results to calculate 10^[3i/8].

For example, for 10^(i/8), I am supposed to get .95995 + .28402i.

Bluntly put, my instructor did not do an example in class.

This is what I do understand from class.
I understand how to use Briggs table to solve log(1.2).
N=1.2
1.2=n1n2n3...
N>n1
n1 found in Briggs table (2nd column--10^x (N), the result closest equal or less than N)

So N/n1=1.2/1.15478 = 1.039159.

n2 is found from the same column on Briggs Table, using 1.039159 (the result from above)

So you get:
N/(n1*n2) = 1.2/(1.15478*1.03663)=1.0024356

and you continue on until you go off Briggs table and then you use the formula
N=n1n2n3(1+deltaN)

s=small number
s=log(1+2.3026s)
DeltaN=.0001852 (I only calculated n1, n2, n3 and used the last result as DeltaN)
DeltaN=2.3026s
so
s=deltaN/2.3026=8.04 * 10^-5.

log (1.2) = n1+n2+n3+8.04*10^-5.

So how do I use this lecture to help me solve the above problem? I just need a start. I feel like I am not doing the Log(10^(i/8)), but rather I have to work backwards, but I don't know how.

Thanks!
Jaden


Answer
Questioner:   Jaden
Category:  Advanced Math
Private:  No
 
Subject:  How to start the problem
Question:  Hello!

Ok, so this is what I have to do. I have to make a chart of the real and imaginary numbers and graph them on the basis of the results I obtain from these problems. I have to calculate 10^[i{(1/2)^10}], 10^[i{(1/2)^9}],10^[i{(1/2)^8}] and so forth until 1/2^0. Then I must calculate 10^[i/8], then 10^[i/4], and use those results to calculate 10^[3i/8].

For example, for 10^(i/8), I am supposed to get .95995 + .28402i.

Bluntly put, my instructor did not do an example in class.

This is what I do understand from class.
I understand how to use Briggs table to solve log(1.2).

>> Use proper vocabulary. (It will definitely raise your own understanding level.)  You don't solve  log(1.2), you compute it.


N=1.2
1.2=n1n2n3...
N>n1
n1 found in Briggs table (2nd column--10^x (N), the result closest equal or less than N)

So N/n1=1.2/1.15478 = 1.039159.

n2 is found from the same column on Briggs Table, using 1.039159 (the result from above)

So you get:
N/(n1*n2) = 1.2/(1.15478*1.03663)=1.0024356

and you continue on until you go off Briggs table and then you use the formula
N=n1n2n3(1+deltaN)

s=small number
s=log(1+2.3026s)
DeltaN=.0001852 (I only calculated n1, n2, n3 and used the last result as DeltaN)
DeltaN=2.3026s
so
s=deltaN/2.3026=8.04 * 10^-5.

log (1.2) = n1+n2+n3+8.04*10^-5.

So how do I use this lecture to help me solve the above problem? I just need a start. I feel like I am not doing the Log(10^(i/8)), but rather I have to work backwards, but I don't know how.

Thanks!
Jaden
........................................................
Hi, Jaden,

A lot of your question is incoherent, I am afraid.  But you certainly tried hard to send me your work, so I'll see what I can do.  

'Briggs Table' is a phrase that has not been used since the 1960's, but I was alive then, so I think it means the Table of Common Logarithms.

And I don't think you need it for these examples.  Your Windows calculator should work fine.

In these examples:

10^[i{(1/2)^10}] = 10^[i{1/1024}],
10^[i{(1/2)^9}] = 10^[i{1/512}],
10^[i{(1/2)^8}] = 10^[i{1/256}]
.....
10^[i{(1/2)^0}] = 10^[i]

you can use these facts:

I)  e^(ix) = cos x + i sin x  << definition attributed to Euler.

II)  10 = e^(ln 10)

So your general case is:

 10^[i{(1/2)^n}] =

 e^(ln 10)^[i{(1/2)^n}] =

 e^[(ln 10)[i{(1/2)^n}] =

 e^[i ln 10{(1/2)^n}]

Now that is in your e^ix form, so it is:

cos (ln 10{(1/2)^n})  +  i sin (ln 10{(1/2)^n})

Now for, say,  n = 3, the example you gave:

ln 10{(1/2)^3} = ln 10(1/8)

On your Windows calc:
I. Choose scientific view.
II. Click Radians button

Enter 10
Press ln button.
Press  '/' (divide)
Enter 8
Press '=' (equals)
(You have 0.28782313662425571050224893183555)
Press MS (memory save)

Press COS.  

You have 0.95886408351651849420869494421575

Press MR (memory recall.)

You get back  0.28782313662425571050224893183555

You get  0.28386558322915274846259802899049

The answer is :

0.95886408351651849420869494421575 + 0.28386558322915274846259802899049i

which you can round off a bit.
.............................

and so forth until 1/2^0, which equals 1, of course.
.........................
Then I must calculate 10^[i/8], then 10^[i/4], and use those results to calculate 10^[3i/8].

>> Of course, you can use the above method for 10^[3i/8], or take

10^[i/8] TIMES 10^[i/4]

which will require a bit of computation:

If  10^[i/8] = a + bi,
and 10^[i/4] = c + di.

Then 10^[3i/8] = (ac - bd) + (ad + bc)i, and you can use YOUR calculator for that.