# Calculating inverse of $(7+9i)$

How can I calculate $(7+9i)^{-1}$?

So I have:

$(7+9i)^{-1}$

$(a+bi) \cdot (7+9i)$

$7a + 9ai + 7 bi + 9bi^2 = 1 + 0i$

$7a + (9a + 7b)i - 9b = 1$

So there are two equations:

1) $7a - 9b = 1$

2) $9a + 7b = 0$

So getting a from the first equation:

$a = \frac{9}{7}b$

Inserting it in the second one:

$9 \cdot \frac{9}{7}b + 7b = 0$

$\frac{81}{7}b + 7b = 0$

$b = \frac{130}{7}$

The correct solution should be: $\frac{7}{130}- \frac{9i}{130}$

Question: My solution looks close but wrong is wrong. Where is my mistake here?

• Great how everyone first answered and then read the question. But good edits though, Commented Apr 2, 2017 at 15:57
• @Ennar I guess: why bother patching up an inefficient method when there are both better tricks and direct formulae?
– user228113
Commented Apr 2, 2017 at 15:59
• @G. Sassatelli, well, obviously to help OP who asked the question to see what went wrong with their method. You should suggest better approach after that, otherwise why not just close this as a duplicate? Commented Apr 2, 2017 at 16:02
• All I'm saying is that we insist on showing effort before asking a question and otherwise close questions as duplicates at best (and even if the question is not a duplicate), but when someone shows effort we are not going to bother to read it? Then why are we asking for that in the first place? To feel better about ourselves for solving their homework? I'm not sure. Commented Apr 2, 2017 at 16:11

Hint:

$z\cdot \bar z=|z|^2$ so $z\cdot \frac{\bar z}{|z|^2}=1$

from the equation 1) $a=\frac{1+9b}{7}$

From the first equation you get $7a = 1 +9b$, not $7a = 9b$.

I think it's easier to use the "multiply by the conjugate" method:

$$\frac1{7+9i} = \frac1{7+9i} \cdot \frac{7-9i}{7-9i} = \cdots$$

This is more of a standard method and when done correctly it guarantees that the denominator is real.

But if for some reason you need to do it the way you started, then from $7a - 9b = 1$, you don't get $a = \frac97 b$. But you do get $a = -\frac79b$ from the second equation.

$$(7+9i)^{-1}=\frac{1}{7+9i}=\frac{1}{7+9i} \times \frac{7-9i}{7-9i}=\frac{7-9i}{7^2-(9i)^2}=\frac{7-9i}{130}= \frac{7}{130}-\frac{9i}{130}$$

Also in your solution $7a-9b=1 \implies a= \dfrac{1+9b}{7} \neq \dfrac{9b}{7}$