Finding Real Numbers for Complex Number Division

[Saladsamurai]

Suppose that a and b are real numbers, not both 0. Find real numbers c and d such that

\frac{1}{a+bi}=c+di

What do they mean find c & d? That's two unknowns is it not?

Anyway. My attempt:

\begin{array} \frac{1}/({a+bi})=c+di\\ \Rightarrow (a+bi)(c+di)=1\\\Rightarrow ac-bd+(ad+bc)i=1\\\Rightarrow \frac{1+bd-ac}{ad+bc}=i\end{array}

Now I know that I could square both sides to get rid of i, but I am not sure how whether that will help or hurt.

 

[Dick]

Squaring won't get rid of the i. Use complex conjugates. Multiply the left side by (a-bi)/(a-bi). Equate real and imaginary parts.

 

[Saladsamurai]

Squaring won't get rid of the i. Use complex conjugates. Multiply the left side by (a-bi)/(a-bi). Equate real and imaginary parts.

Okay. But why won't squaring both sides of my last line get rid of i? I mean, it will leave me dead in the water, but the RHS will surely be -1 right? (Just want to be sure i am not missing something big here)

 

[Saladsamurai]

If I multiply the original EQ by the conjugate I get

\frac{a-bi}{a^2+b^2}=c+di

I am not sure what you mean by equate the real and imaginary parts? Sorry, I have never really used complex numbers

on the LHS, I have the complex number \frac{a}{a^2+b^2}+\frac{-b}{a^2+b^2}i

Does that form help my plight?

 

[Dick]

Sure. The real part of the left side is a/(a^2+b^2), the real part of the right side is c. So c=a/(a^2+b^2). (a,b,c,d are REAL). Now equate the imaginary parts of both sides. You are supposed to find c and d in terms of a and b.

 

[Saladsamurai]

Ohhhhh. i see where my confusion lies. a and be are real numbers, BUT we refer to the quantity (b*i) as the imaginary PART right?

i is the only imaginary number but b*i is the imaginary part

 

[Cyosis]

If you have a complex number a+ib, then a is the real part and b is the imaginary part not ib.

 

[Mark44]

In other words, the imaginary part is the coefficient of i. That means that both the real part and the imaginary part are real numbers.

 

[HallsofIvy]

and bi is an imaginary number, not only i.

 

[HallsofIvy]

Unfortunately, you can find as many authoritative statements that the "imaginary part" of a+ bi is "bi" as you can that it is "b". Fortunately, it is not a critical distinction.
If
\frac{a- bi}{a^2+ b^2}= \frac{a}{a^2+ b^2}- \frac{bi}{a^2+ b^2}= c+ di
then
\frac{a}{a^2+ b^2}= c
and
\frac{-b}{a^2+ b^2}= d
no matter what you call them!