What is the multiplicative inverse of -9-17i?

[GrannySmith]

What is the multiplicative inverse of -9-17i? Would it be -9+17i?

For 12 it would be 1/12 right? What would it be for a complex number? 1/(-9 - 17i)? I'm so confused lol.

 

[daon2]

What is the multiplicative inverse of -9-17i? Would it be -9+17i?

Is this a lack of motivation today or just not knowing what "multiplicative inverse" is? You should be able to answer that in a few seconds with a simple multiplication.

For 12 it would be 1/12 right? What would it be for a complex number? 1/(-9 - 17i)? I'm so confused lol.

Yes, if \(\displaystyle z\neq 0\) then the multiplicative inverse of \(\displaystyle z\) is \(\displaystyle \dfrac{1}{z}\). I suspect you are required to write it as a sum of its real and imaginary parts.

I'm sure this is in your text, but try multiplying \(\displaystyle \dfrac{1}{x+iy}\) by \(\displaystyle \dfrac{x-iy}{x-iy}\) and see what happens.

 

[DrPhil]

What is the multiplicative inverse of -9-17i? Would it be -9+17i?

For 12 it would be 1/12 right? What would it be for a complex number? 1/(-9 - 17i)? I'm so confused lol.

Another was to write it without using "division" is that you want the number that results in 1 (or 1+0i) when you multiply by your given number

12* (1/12) = 1


(-9 - 17i) * (x + iy) = (1 + 0i)

Multiply the two binomials using FOIL and separate real and imaginary parts to get two equations in two unknowns.

 

[pka]

What is the multiplicative inverse of -9-17i? Would it be -9+17i?

For 12 it would be 1/12 right? What would it be for a complex number? 1/(-9 - 17i)? I'm so confused lol.

Here is a useful formula: \(\displaystyle \displaystyle\dfrac{1}{z} = \dfrac{{\overline z }}{{{{\left| z \right|}^2}}}\)