Complex Number help :) (1 Viewer)

[missAy]

Hey BoSer's I'm having some trouble with complex numbers, any help would be greatly appreciated

4. b) by expressing 1+ (sq-root3)i in mod-arg form, show that z = (sq-root8) cis (7π)/12


that's the first part of the question which I can do, here's where I'm lost

c) Hence, find the exact values of cos (7π)/12 and sine (7π)/12


Also:
6. If z=cis q, prove the following expressions
a) (1-i)z = (sq-root2)cis [q - (π/4)]

b)[-1+i(sq-root3)]/z = 2cis[(2π/3) - q]

Thanks

 

[jyu]

Hey BoSer's I'm having some trouble with complex numbers, any help would be greatly appreciated

4. b) by expressing 1+ (sq-root3)i in mod-arg form, show that z = (sq-root8) cis (7π)/12


that's the first part of the question which I can do, here's where I'm lost

c) Hence, find the exact values of cos (7π)/12 and sine (7π)/12


Also:
6. If z=cis q, prove the following expressions
a) (1-i)z = (sq-root2)cis [q - (π/4)]

b)[-1+i(sq-root3)]/z = 2cis[(2π/3) - q]

Thanks

What is in 4a) ?

For 6)

Change 1- i , -1 + sqrt3 i to polar form. Then multiply and divide in polar form.

 

[missAy]

What is in 4a) ?

If z = [1+(sq-root3)i] x (1+i)

4a) Express z in the form x+iy, where x and y are real

 

[jyu]

If z = [1+(sq-root3)i] x (1+i)

4a) Express z in the form x+iy, where x and y are real

Work out z in polar form to be 2sqrt(2) cis(7pi/12).

.: (1-sqrt(3)) + (1+sqrt(3)i = 2sqrt(2) cis(7pi/12)

.: [(1-sqrt(3)) + (1+sqrt(3)i]/2sqrt(2) = cis(7pi/12)

.: [(1-sqrt(3)) + (1+sqrt(3)i]/2sqrt(2) = cos(7pi/12) + isin(7pi/12)

Compare the two sides to obtain answers.

:wave:

 

[missAy]

Work out z in polar form to be 2sqrt(2) cis(7pi/12).

.: (1-sqrt(3)) + (1+sqrt(3)i = 2sqrt(2) cis(7pi/12)

.: [(1-sqrt(3)) + (1+sqrt(3)i]/2sqrt(2) = cis(7pi/12)

.: [(1-sqrt(3)) + (1+sqrt(3)i]/2sqrt(2) = cos(7pi/12) + isin(7pi/12)

Compare the two sides to obtain answers.

:wave:

Thanks but I did a) and b) but I didn't understand c)

Well the answer for 4c is

Cos (7π)/12 = [1+(sq-root3)]/(sq-root8)

Sine (7π)/12 = [1+(sq-root3)]/(sq-root8)

I feel like I'm missing something really obvious, coz they're conjugate pairs or something

 

[jyu]

Thanks but I did a) and b) but I didn't understand c)

Well the answer for 4c is

Cos (7π)/12 = [1+i(sq-root3)/(sq-root8)

Sine (7π)/12 = [1+i(sq-root3)]/(sq-root8)

I feel like I'm missing something really obvious, coz they're conjugate pairs or something

Continue from my last post.

[(1-sqrt(3)) + (1+sqrt(3)i]/2sqrt(2) = cos(7pi/12) + isin(7pi/12)

.: (1-sqrt(3))/2sqrt(2) + (1+sqrt(3))i/2sqrt(2) = cos(7pi/12) + isin(7pi/12)

.: cos(7pi/12) = (1-sqrt(3))/2sqrt(2) and sin(7pi/12) = (1+sqrt(3))/2sqrt(2)

Note: 2sqrt(2) = sqrt(8)
For cos(7pi/12), your book answer and my answer differ by +/-. Check the working and see which is correct.

:wave:

By the way, cos@ = sin@ only when @ = pi/4 +/- n(pi), where n = 0, 1, 2, ...

 

[toadstooltown]

To be a lot clearer,
a)z=1-√3+i(1+√3)

b)z=(1-i√3)×(1+i)
=2cis[pi/3]×√2cis[pi/4]
=2√2cis[7pi/12]
=√8cis[7pi/12]

c)so z=1-√3+i(1+√3)
and z=√8cis[7pi/12]
1-√3+i(1+√3)=√8cis[7pi/12]
1-√3+i(1+√3)=√8cos[7pi/12]+√8sin[7pi/12]
We know equate real and imaginary parts, as for a+ib to be equal to x+iy a=x, b=y
1-√3=√8cos[7pi/12],1+√3=V8sin[7pi/12]
cos[7pi/12]=(1-√3)/(2√2),
sin[7pi/12]=(1+√3)/(2√2)


"Well the answer for 4c is

Cos (7π)/12 = [1+i(sq-root3)/(sq-root8)

Sine (7π)/12 = [1+i(sq-root3)]/(sq-root8)"

this is wrong, this is somewhere along the way to equating real and imaginary parts, no i's should be in it. trig{ax}, a is real, has the answer real. So the above statement can't ever be true.

 

[missAy]

this is wrong, this is somewhere along the way to equating real and imaginary parts, no i's should be in it. trig{ax}, a is real, has the answer real. So the above statement can't ever be true.

Oh gosh ur right, sorry, I'll be more careful next time and thanks.

 

[missAy]

Everyone on BoS is so helpful

Anyway cuz I can't seem to edit my post, I can do 6 a) and b) but I got really stuck on this one.

Where z=cisq, prove the following:
d) -(1+i) (( √3) + i) z = 2(√2) [cis (q - (7π/12))]

(the spaces mean multiply)<O</O

 

[Slidey]

Ha! This was a bit nasty until I converted the negative sign at the front to cis(pi) or cis(-pi).

LHS = cis(-pi).sqrt2.cis(pi/4).2.cis(pi/6).cisq, yes?
LHS = sqrt(8).cis(q-pi+5pi/12)
LHS = sqrt(8).cis(q-7pi/12)

Voila! (sqrt8 is of course 2sqrt2)

EDIT: Was this a "prove using steps x and y" question? I guess that's what zeek did.

 

[zeek]

d) -(1+i)(sqrt(3)+i)z=2sqrt(2)[cis(@-7pi/12)]

LHS
======

Multiply the two individual complex numerals together to make them one...

[(1-sqrt(3))-(1+sqrt(3))i]z ----> Transform this into Mod-Arg Form:

@ = tan^-1{[-1-sqrt(3)]/[1-sqrt(3)]} {after rationalising}
= tan^-1{sqrt(3)+2)
= 5pi/12 {in radians}

R = sqrt{[1-sqrt(3)]² + [1+sqrt(3)]²}
= sqrt(8)

.: LHS = [sqrt(8)cis(5pi/12)]z

Now, since the argument in Mod-Arg form must be -pi<@<=pi, the current argument (5pi/12) can be changed to (-7pi/12). Look at the attachment for a drawing of the angles.

.: LHS = 2sqrt(2)cis(-7pi/12) . z {z=cos@}
= 2sqrt(2)cis(@-7pi/12)