Need some help with Complex Number Q. (2 Viewers)

[peterpandied]

Solve z^2 + (1+i)z + 2i = 0, expressing the roots in mod-arg form. If z1, z2 are these roots, prove that |z1| = |z2| = /2 (uh how do you do square root signs..)
, and that arg z1 + arg z2 = pie/2

Well this question is really confusing me.. and seems like I m not getting any where..
please help~!

 

[mitsui]

Hmm methods:

1.use quadratic formula to find the roots. if u come across a unreal square root, use z=x+iy to solve it.

then u should able to get two roots for that eqution

then u find the mod, and u should find that they r all equals to /2
and find arg, and they should add up to 90degrees

i am not good wif doing it on comp. hope it helps a bit at least

 

[Riviet]

Solve z^2 + (1+i)z + 2i = 0, expressing the roots in mod-arg form. If z1, z2 are these roots, prove that |z1| = |z2| = /2 (uh how do you do square root signs..)
, and that arg z1 + arg z2 = pie/2

Well this question is really confusing me.. and seems like I m not getting any where..
please help~!

As mitsui stated, use the quadratic formula.

z={-(1+i)+sqrt[(1+i)²-8i]} / 2

z=[-1-i+sqrt(-6i)] / 2

Now let sqrt(-6i)=a+ib

-6i=a²-b²+2abi

equating real and imaginary parts,

a²-b²=0 *

ab=-3 **

Solving * and ** simultaneously,

a=+sqrt3

.: b=+sqrt3

.: sqrt(-6i)=sqrt3 - (sqrt3)i, -sqrt3 + (sqrt3)i

so z= [-1-i+sqrt3-(sqrt3)i]/2 or [-1-i-sqrt3+(sqrt3)i]/2

z₁=[-1+sqrt3-(1+sqrt3)i]/2, z₂=[-1+sqrt3-(1+sqrt3)i]/2

Now use |z|=sqrt(x²+y²) where x and y are the real and imaginary parts of z₁ and z₂ and show they both equal sqrt2.

Then find the arg of each root by using tan@=y/x. Add these two arguments and show it equals 90 degrees or pi/2 radians.

I hope that helps,

Riv

 

[KeypadSDM]

z₁ + z₂ = - 1 - i
z₁ * z₂ = 2i

Thus

arg(z₁) + arg(z₂) = arg(z₁ * z₂) = arg(2i) = pi/2

I don't appreciate the question's need to express them in mod-arg form, but if we must:

z^2 + (1+i)z + 2i = 0
using the quadratic formula:
z = (-b + Sqrt[b² - 4ac])/(2a)
where a = 1, b = 1 + i, c = 2i

Also, you might notice that b² = c

Thus:

2z = - 1 - i + Sqrt[(1 + i)² - 8i]

Note: Sqrt[(1 + i)² - 8i] = Sqrt[2i - 8i] = Sqrt[-6i]

-6i = 6cis[-pi/2]
Sqrt[-6i] = Sqrt[6]cis[-pi/4] = Sqrt[3] * Sqrt[2]cis[-pi/4] = Sqrt[3] * (1 - i)

So we have:

2z = - 1 - i + Sqrt[3] * (1 - i)
= - 1 - i + (-i)Sqrt[3] * (i + 1) [multiplying the sqrt by -i, and the inside of the bracket by i]
= (1 + i)(-1 + (-i)Sqrt[3])
=-(1 + i)(1 + iSqrt[3])

2z = cis(pi) * Sqrt[2]cis(pi/4) * 2 * cis(+ pi/3)
z = Sqrt[2] * cis(5pi/4 + pi/3)

Thus:

z₁ = Sqrt[2] * cis(5pi/4 + pi/3) = Sqrt[2] * cis(19pi/12)
z₂ = Sqrt[2] * cis(5pi/4 - pi/3) = Sqrt[2] * cis(11pi/12)

EDIT: I found the problem, the arg of 1 + i is pi/4 not pi/2

Thus the args add to:
30pi/12 = 2pi + pi/2 [as required]

And the moduli of both are Sqrt[2]

 

[icycloud]

Edit Keypad found a mistake in my working (see below), so just use his solutions above .

 

[KeypadSDM]

P(z) = z^2 + (1+i)z + 2i = 0

Using sum of roots of polynomials:

z_1 + z_2 = -b/a = -1-i
|z_1 + z_2| = |z_1| + |z_2|
= |-1-i|
= Sqrt[1 + 1] = Sqrt[2]

Yeah, dumbest algebra ever.

Set a = i, b = -i
|a + b| = 0
|a| + |b| = 2

 

[mitsui]

oh wow. smart. O_O

 

[icycloud]

Yeah, dumbest algebra ever.

Set a = i, b = -i
|a + b| = 0
|a| + |b| = 2

Oops sorry, got confused with conjugates. Funny how it works for that question though. Will edit my post now.

 

[peterpandied]

Thx guys~!