Question

What is the real and imaginary parts of the complex number -5+6i?

Answer 1

A complex number is written in the form of a+bi, where a is the real part and b is the imaginary part.

Answer 2

Can you also help me with multiplying complex numbers? I am not good with it. For example, i(2+i) equals what? And (1+2i)-(1+5i) equals what? And then there's simplifying them.
Grrrrrrr.

Answer 3

To multiply (2+3i) with (5-2i) you do:

\[(2+3i)*(5-2i) = 2*5 - 2*2i + 3i*5 - 2i*3i\]

Then you group and replace i^2 with -1.

The subtraction is done in the same way with addition. In general, you treat i as a constant. The difference is that it's square equals -1.

Answer 4

So i equals to a number that's actually not a number? So, how do you simplify complex numbers then? \[\frac{ 5+2i }{ 6+i }\]

Answer 5

i is the imaginary unit. I told you to treat it as a constant, so you will multiply (2+3i) with (5-2i) as if you had (2+3a) with (5-2a), where a is a constant.

Answer 6

About the division
\[\frac{ 5+2i }{ 6+i }\]
you want to get rid of the 6+i. You want to make it real. So what is the result if you multiply (6+i) with (6-i)?

Answer 7

Oh, oh! I get it now! Oh my goodness, thank you! Yes!

Answer 8

Did you find the result of the multiplication (6+i) with (6-i) ?

Answer 9

Is it 37?

Answer 10

That's right! Bravo! Can you see now what happens to
\[\frac{ 5+2i }{ 6+i }\]
when you multiply both numerator and denominator with (6-i)?