Question

What is the additive inverse of the complex number 9 – 4i?

Answer 1

Additive inverse should get you to zero when you apply it:\[\large\rm 9-4i+z=0\]

Answer 2

Where z is some complex number, x+iy,
\[\large\rm 9-4i+x+iy=0\]Equate the real parts,\[\large\rm 9+x=0\]Solve for x,
and equate the imaginary parts,\[\large\rm -4i+yi=0\]and solve for y.

Answer 3

so –9 + 4i

Answer 4

@zepdrix

Answer 5

yay good job \c:/

Answer 6

what about
What is the value of the product (3 – 2i)(3 + 2i)?

Answer 7

5
9 + 4i
9 – 4i
13

Answer 8

@zepdrix

Answer 9

You can FOIL it out,
\(\rm (a+bi)(a-bi)=a^2+abi-abi-b^2i^2\)
But what you'll find is that the middle terms always cancel out,
and what you're left with,
\(\rm =a^2+b^2\)
is the sum of squares.

This only applies when the brackets are exactly the same,
but with a different sign between the terms.
we call these conjugates.

So therefore, by that rule:
\(\rm (3-2i)(3+2i)=3^2+2^2\)

Answer 10

5

Answer 11

@zepdrix

Answer 12

3^2+2^2 is 5? Hmm

Answer 13

9 + 4i