1. ## Beginners Complex Analysis

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

(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!

2. What did you do?
These look straightforward.

3. 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.

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 $z=\pm 6i$.

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

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

(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.

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 ?

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...