Hello again! A few complex no. questions have got me stuck.

If z = 1 + i*root3

Find z bar to the power of -5.

If w = root2cis(pie/4)

Find w/z in polar and cartesian form and thus deduce an exact value for cos(pie/12).

Printable View

- Feb 16th 2007, 01:02 AMclassicstringsComplex Numbers polar
Hello again! A few complex no. questions have got me stuck.

If z = 1 + i*root3

Find z bar to the power of -5.

If w = root2cis(pie/4)

Find w/z in polar and cartesian form and thus deduce an exact value for cos(pie/12). - Feb 16th 2007, 03:55 AMAlvinCY

z = 1 + root 3 i

z bar = 1 - root3 i

Converting this to cis form, z bar = 2 cis (-pi/3)

Using De Moivre's Theorem, (z bar)^5 = (2^5) cis 5*(-pi/3) = 32 cis (pi/3)

If w = root 2 cis (pi/4)

w/z in polar form would be (32/ root 2) cis (pi/3 - pi/4) = 16 root 2 cis (pi/12)

w/z in cartesian form would be:

w in cartesian form is root2 (cos pi/4 + i sin pi/4) = root 2 [(1/root 2) + i (1/root 2)] = 1 + i

w/z = (1+i)/(1 + root 3 i)

Realising the denominator would give you w/z = (1 + root 3)/4 + (1 - root 3)/4 * i

and thus deduce an exact value for cos(pi/12).

We know that 16 root 2 cis (pi/12) = 16 root 2 [cos (pi/12) + i sin (pi/12)] = (1 + root 3)/4 + (1 - root 3)/4 * i

So 16 root 2 * cos (pi/12) = (1 + root 3)/4

cos (pi/12) = (1 + root 3)/4 divided by 16 root 2...

Simplify and realise the answer.

Done! - Feb 20th 2007, 12:59 AMclassicstrings
I worked through the problem and still had trouble with the last part.

I was wondering if someone can check my solutions and perhaps give me a faster more logical way of working through this question. Thanks.

If z = 1 + i*root3

i) Find the modulus and argument of z

ii) Express z^5 in Cartesian form a + ib where a and b are real

iii) Find z*zbar

iv) hence, find zbar to the power of negative 5

If w = root2 cis (π/4)

i) Find w/z in polar form.

ii) Express z and w in cartesian form and hence find w/z in cartesian form.

iii) Use answers from i) and ii) to deduce exact value for cos(π/12)

The attempt at a solution

i) Modulus is 2, Argument is π/3

ii) Using De Moivre's Theorem, 2^5cis(5*(π/3))

= 32cis(-pie/3)

Then changing to Cartesian form - 32 (cos(-π/3) + isin(-π/3))

= 32(0.5 + (root3/2)i)

= 16 + 16i*root3

iii) z = 1 + i*root3, zbar = 1 - i*root3

z*zbar = 4

iv) Now this part I don't understand the "hence", as in how am I meant to use previous results to get this answer?

I try - zbar = 2cis(-π/3)

Then zbar^-5 = Using DM Theorem, (1/32)cis(5π/3)

= (1/32)cis(-π/3)

Then convert to cartesian - (1/32)(cos(-π/3) + isin(-π/3))

= (1/32)(0.5 + iroot3/2)

= 1/64 + iroot3/64

If w = root2 cis (π/4)

i) Find w/z in polar form.

I found it in cartesian - (1+i)/(1 + root3 i) then realising, gives (1 + root3)/4 + (1-root3)/4 * i

But how do I change to Polar form? I am also stuck on how to get cos(pie/12) as exact value, even after reading Alvin's comments. - Feb 20th 2007, 02:03 AMclassicstrings
Ok I managed to get another part.

In polar form, it is

(root2/2)cis(-pie/12)

**Ok after re-reading Alvin's solution I got it myself, thanks!**:D - Feb 20th 2007, 03:51 AMtopsquark
- Feb 20th 2007, 08:54 AMSoroban
Hello, classicstrings!

Both of you forgot the "minus" . . .

Quote:

If .z .= .1 + i√3

Find z-bar to the power of**-5**.

We have: .z .= .2·cis(π/3) .2[cos(π/3) + i·sin(π/3)]

Then: .z-bar .= .2[cos(-π/3) + i·sin(-π/3)] .= .2[cos(π/3) - i·sin(π/3)]

Then: .(z-bar)^{-5} .= .(2^{-5})[cos(-5π/3) - i·sin(-5π/3)]

. . . . . . . . . . . . . . . = .(1/32)[(1/2) - i(√3/2)]

. . . . . . . . . . . . . . . = .(1/64)(1 - i√3)