Solutions to this equation - complex numbers

**1.**

The question says, what is the number of solutions to this equation in the field of complex numbers?

The choices are

a) 1

b) 2

c) 3

d) infinitely many

I have no clue to this one.

**2.** And here's another one:

For any real number x, is equal to

a) +x

b) -x

c) complex

d) pure imaginary

Writing it in the form , we can simplify it to

Which shows that the number is not real (since no number to the power four would give -x). So +x and -x are eliminated. But what about the last two choices? What is the difference between complex and pure imaginary?

Re: Solutions to this equation - complex numbers

2) You can write:

What about ?

You know every complex number has a real part and an imaginary part, if the real part is equal to zero (so if there's no real part, only an imaginary part) then we call it pure imaginary.

Re: Solutions to this equation - complex numbers

Quote:

Originally Posted by

**Siron** 2) You can write:

What about

?

You know every complex number has a real part and an imaginary part, if the real part is equal to zero (so if there's no real part, only an imaginary part) then we call it pure imaginary.

Thanks for the reply!!

Complex numbers are of the form a + b i (plus sign in between the complex and imaginary part). The number has a multiply sign in between so even if is real, this does not qualify as a complex. Does this mean that this would be pure imaginary?

Also, any clue about the first question?

Re: Solutions to this equation - complex numbers

Every imaginary number **is** a complex number: 0+ bi.

Re: Solutions to this equation - complex numbers

2. As HallsofIvy has already pointed out, the imaginary numbers are in fact a subset of the complex numbers.

We can easily rewrite this expression as the following:

Sadly I was unable to come up with a solution of my own after two attempts, however a nice solution is presented here.

Your answer is C.