Complex q's (1 Viewer)

x.Exhaust.x · Jan 31, 2009

1. Is it possible to find the conjugate of z=10i-10/-8-4i? I'm stuck on this part of my 1 paged working lol.

2a) Find all integers x and y such that (x+iy)^2=-3-4i where x and y are real numbers.

I've got y=+-2, x=-+1. Is this right?

b) Hence, or otherwise, solve the equation z^2-7z+13+i=0

3a) If z=x+iy, determine the values of Re(z^2-3z) and Im(z^2-3z).

Is it simply Re(z^2-3z)=x^2-3x and Im(z^2-3z)=y^2-3y? I don't think it would be that simple...

b) If z=1-i/2+1, show that Re(z+z^-1)=7/10

4a) Evaluate:

Arg[cis7pie/5], Arg[9pie/5]

Thanks

.

independantz · Jan 31, 2009

x.Exhaust.x said:
1. Is it possible to find the conjugate of z=10i-10/-8-4i? I'm stuck on this part of my 1 paged working lol.

2a) Find all integers x and y such that (x+iy)^2=-3-4i where x and y are real numbers.

I've got y=+-2, x=-+1. Is this right?

b) Hence, or otherwise, solve the equation z^2-7z+13+i=0

3a) If z=x+iy, determine the values of Re(z^2-3z) and Im(z^2-3z).

Is it simply Re(z^2-3z)=x^2-3x and Im(z^2-3z)=y^2-3y? I don't think it would be that simple...

b) If z=1-i/2+1, show that Re(z+z^-1)=7/10

4a) Evaluate:

Arg[cis7pie/5], Arg[9pie/5]

Thanks .

1)conjugate of z= conjugate of {10i-10/-8-4i]=conjugate of (10i-10)/conjugate of [-8-4i]- this is a rule

therefore, conjugate of z=-10-10i/-8+4i
=(-10-10i)(-8-4i)/(-8+4i)(-8-4i)
=40+120i/80
=1/2+3/2i

2) a.
$(x+iy)^2=-3-4i \text{ ,where x and y are real}\\x^2+2xyi-y^2=-3-4i\\x^2-y^2=-3 2xy=-4, \text{ equating real and imaginary parts}\\x^2-y^2=-3-[1] y=-2/x-[2]\\\text{from 1 and 2 we get:}\\x^2-(-2/x)^2=-3\\x^4+3x^2-4=0\\(x^2+4)(x^2-1)=0 \text{ but x is real}\\i.e. x=\pm 1 -[3]\\\text{ from 2 and 3 we get:}\\\text{when} x=1, y =-2\\\text{when} x=-1, y =2\\$

b.
$z^2-7z+13+i=0\\ z=[7\pm \sqrt{49-4(1)(13+i)}]/2\\ z=[7\pm \sqrt{-3-4i}]/2\\ z=[7\pm 1-2i]/2 \text{ Using Part a}\\ z=7+1-2i/2 \text{ Or } z=7-(1-2i)/2\\ z=8-2i/2 \text{ Or } z=6+2i/2\\ z=4-i \text{ Or } z=3+i\\$

3)
a.
$z^2-3z=(x+iy)^2-3(x+iy)=x^2+2xyi-y^2-3x-3yi\\ \text{therfore: } re(z^2-3x)=x^2-y^2-3x\\ \text{therfore: } im(z^2-3x)=2xy-3y\\$

b.
$z+z^-1= [1-i/2+i]+[1-i/2+i]^-1\\ =[1-i/2+i]+[2+i/1-i]\\ =[(1-i)^2+(2+i)^2/(2+i)(1-i)]\\ =3+2i/3-i\\ =(3+2i)(3+i)/(3-i)(3+i)\\ =7+9i/10\\ =7/10+9i/10\\ \text{therefore: } Re(z+z^-1)=7/10\\$

x.Exhaust.x · Jan 31, 2009

Thankz independentz!

Complex q's (1 Viewer)

x.Exhaust.x

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