Hence express tan (pi/12) in surd form.

[charmedbeauty]

Homework Statement

Let z= √3 -i

and w= 1+i

i)calculate zw in "a+ib"form

ii)Write z and w in polar form and thus write zw in polar form

iii)Hence express tan (pi/12) in surd form.

Homework Equations

The Attempt at a Solution

(√3 -i)(1+i)

= (√3+1)+(√3i-i)

ii)

|z|=2 → Arg(z)=tan^-1(√3/-1) = -pi/3

so z = 2*e^-i(pi/3)

|w|= √2 → Arg(w)=pi/4

so w=√2 *e^i(pi/4)

so zw= 2√2*e^i(pi/4 -pi/3)

= √8*e^-i(pi/12)


iii) this is the part I'm having trouble with...

so I think I should convert it into "a +ib" form

so |zw| = √8 → Arg(zw)=-pi/12

but then I am stuck because there is no exact result for -pi/12

?

 

[tiny-tim]

hi charmedbeauty!

(have a pi: π and try using the X² button just above the Reply box )

hint: if e^iπ/12 = p + iq, what is tan(π/12) ?

 

[Curious3141]

Arg(z)=tan^-1(√3/-1) = -pi/3

so z = 2*e^-i(pi/3)

Check the math here very carefully, then redo what you did, then apply TinyTim's hint.

 

[charmedbeauty]

Check the math here very carefully, then redo what you did, then apply TinyTim's hint.

Im pretty sure the math here checks out?

unless it should be 2e^i(-π/3)

but I am still stuck on tinytims hint

I know that tan =sin/cos

and 2√2 e^i(π/12)

= 2√2[cos(π/12)+isin(π/12)]

but I am stumped to see how you can relate these?

and as I understand it I shouldn't be trying to find the real and imaginary part of Arg(z)=π/12 and mod of √8 because that is not going to turn out nice.

unless I divided the terms by cos(π/12) to get a term with tan but I still think that's wrong?

 

[NewtonianAlch]

\theta=\arctan \left( {\frac {q}{p}} \right), so \tan \left( \theta \right) ={\frac {q}{p}}

theta = Pi/12

You already have q and p.

The other (longer) method is to use the identity (sin x)/(cos x) = tan x

Since Pi/12 = 15 degrees. (sin 45 - 30) = sin(45)cos(30)...etc.etc. substitute in the exact values in surd form, and then divide. The answer works out to the same.

 

[Curious3141]

Im pretty sure the math here checks out?

unless it should be 2e^i(-π/3)

You're supposed to take the arctangent of the Imaginary part/Real part.

The imaginary part here is -1, the real part is √3.

So the argument is arctan (-1/√3) = ?

but I am still stuck on tinytims hint

I know that tan =sin/cos

and 2√2 e^i(π/12)

= 2√2[cos(π/12)+isin(π/12)]

but I am stumped to see how you can relate these?

Right, once you get to this stage, you compare it to the a + bi form you worked out earlier.

So here you're comparing 2√2cos(π/12) + i*2√2sin(π/12) to (√3+1)+i*(√3-1)

Equate the real and imaginary parts.

You can then get a surd expression for the cosine, and another for the sine. Just divide the sine by the cosine (as you know), and simplify (do the usual rationalisation of the denominator), and you're done!

 

[charmedbeauty]

You're supposed to take the arctangent of the Imaginary part/Real part.

The imaginary part here is -1, the real part is √3.

So the argument is arctan (-1/√3) = ?



Right, once you get to this stage, you compare it to the a + bi form you worked out earlier.

So here you're comparing 2√2cos(π/12) + i*2√2sin(π/12) to (√3+1)+i*(√3-1)

Equate the real and imaginary parts.

You can then get a surd expression for the cosine, and another for the sine. Just divide the sine by the cosine (as you know), and simplify (do the usual rationalisation of the denominator), and you're done!

ok thanks I see what you mean about check the math... I should have picked up something was wrong when I had the -π/12 instead of π/12

ok,, I should be right from here thanks for tips, your help really helps!.