Beginners Complex Analysis

shadow_2145 · #1 $Old$ 08-26-2008, 04:45 PM

$z=1+2i, w=2-i, \zeta=4+3i$

(1) Use the quadratic formula to solve these equations; express the answers as complex numbers.

(a) $z^2+36=0$

(b) $2z^2+2z+5=0$

I have more of these but I will try them on my own after I receive help on these.

The number x is called the real part of z and is writtern x = Re z. The number y, despite the fact that it is also a real number, is called the imaginary part of z and is writter y = Im z.

(2) Find Re(1/z) and Im(1/z) if z= x + iy, z $\not=0$ . Show that Re(iz)= -Im z and Im(iz) = Re z.

If further explanation is needed, please let me know. Thanks for the help!

ThePerfectHacker · #2 $Old$ 08-26-2008, 05:34 PM

What did you do?
These look straightforward.

shadow_2145 · #3 $Old$ 08-26-2008, 05:51 PM

Is 1(a)'s answer -4 plus or minus 24i over 72?

Simplified to -1 +- 24i over 18?

If this is right, then I know how to do these, lol.

ThePerfectHacker · #4 $Old$ 08-26-2008, 06:00 PM

Quote:

Originally Posted by shadow_2145 $View Post$

Is 1(a)'s answer -4 plus or minus 24i over 72?

Simplified to -1 +- 24i over 18?

If this is right, then I know how to do these, lol.

This one is $z^2+36=0$ which becomes $z^2 = -36$ .
The solutions are $z=\pm 6i$ .

shadow_2145 · #5 $Old$ 08-26-2008, 06:50 PM

I don't understand how to use the quadratic formula to find these answers. Thanks though.

Moo · #6 $Old$ 08-27-2008, 04:26 AM

Quote:

Originally Posted by shadow_2145 $View Post$

I don't understand how to use the quadratic formula to find these answers. Thanks though.

You know that $i^2=-1$

$z^2+36$ looks like a lot $a^2-b^2=(a-b)(a+b)$ .

So write : $z^2+36=z^2-(-36)=z^2-(6i)^2$ . Here is your quadratic formula

Quote:

(b) $2z^2+2z+5=0$

Just use the discriminant method.

In $ax^2+bx+c=0$ , $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, I'm pretty sure $b^2-4ac<0$ and this is where imaginary part intervenes.

Quote:

The number x is called the real part of z and is writtern x = Re z. The number y, despite the fact that it is also a real number, is called the imaginary part of z and is writter y = Im z.

(2) Find Re(1/z) and Im(1/z) if z= x + iy, z\not=0.

$\frac 1z=\frac 1{x+iy}$

When you have a complex number in a denominator, you often have to multiply it by its conjugate (called $\overline{z}$ ), that is $x-iy$ and by using the quadratic formula $(a-b)(a+b)=a^2-b^2$ , we get $(x+iy)(x-iy)=x^2-(iy)^2=x^2+y^2$ , which is a real number.

So $\frac 1{x+iy}=\frac 1{x+iy} \cdot \frac{x-iy}{x-iy}$

Can you go from here ?

Quote:

Show that Re(iz)= -Im z and Im(iz) = Re z.

What is i*z ?
What do you conclude ?

You gotta be able to do these (in a first time, to "know"), because it's quite a basis of the manipulation of complex numbers...

mr fantastic · #7 $Old$ 08-27-2008, 09:52 AM

Quote:

Originally Posted by shadow_2145 $View Post$

I don't understand how to use the quadratic formula to find these answers. Thanks though.

If you showed all your working I'm sure your problem applying the quadratic formula could be easily resolved.

shadow_2145 · #8 $Old$ 08-28-2008, 12:45 PM

Yeah, I figured out the quadratic formula stuff yesterday. Really simple eh? lol I'm going to take a look at the real and imaginary problem later..

shadow_2145 · #9 $Old$ 08-29-2008, 12:37 AM

The real and imaginary stuff turned out to be quite easy too. Thanks