# Solving a complex equation

1. Jan 27, 2013

### Nikitin

Hey! In an example in my textbook, the complex equation z2 +2i*z-(1+i) = 0

is solved using the quadratic formula, so that it ends up with:

z= -i ± √i = -i ± (1+i)/√2 = .....etc.

Uhm.. I got 2 questions:

1) Why isn't the first solution (-i ± √i) acceptable?
2) How does √i = (1+i)/√2 ? Only thing I know is that the root of two is the length of 1+i, but how does dividing 1+i with its length give √i???

thanks

Last edited: Jan 27, 2013
2. Jan 27, 2013

### jbunniii

Who says it isn't acceptable? It may not be in its simplest form, for someone who doesn't want to see complex numbers inside square roots, but it is perfectly correct. Neither answer is in the standard form $a + bi$ where $a$ and $b$ are real numbers. Normally it's a good idea to express the answer that way unless asked to do otherwise.
It is easy to take square roots of complex numbers if we put them into polar coordinates, such as $re^{i\theta}$. In this case we have $i = 1 \cdot e^{i\pi/2}$. What is the square root of that number?

3. Jan 27, 2013

### Nikitin

Ah, I see. √i = e^i*π/4 or e^i*5π/4, => √i = ± (1+i)/√2.

Thank you!

But can you explain me why it is easier to take square roots if you put the complex number in the re^iθ form? It's the same kind of work as if you have the number in r(cosθ + i*sinθ) form, just with a bit less writing. Right?

4. Jan 27, 2013

### jbunniii

Sure, your form is also polar form, so use whichever you prefer. The key is that you can easily see the magnitude $r$ and the angle $\theta$. I was contrasting these forms with rectangular coordinates, which look like $a + ib$ with $a$ and $b$ real, or with other forms such as $\sqrt{i}$, where there's no obvious method to compute roots.

5. Jan 27, 2013

### Nikitin

Ah, thanks. I was unsure what polar form meant, but i see now. Thx again!