Calculating inverse of $(7+9i)$

Question

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?

Answer 1

Hint:

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

Your mistake:

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

Answer 2

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

Answer 3

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.

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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.

Answer 4

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

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