<s>[INST] 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?
 [/INST] $$(7+9i)^{-1}=\frac{1}{7+9i}=\frac{1}{7+9i} \times \frac{7-9i}{7-9i}=\frac{7-9i}{7^2-(9i)^2}=\frac{7-9i}{130}= \frac{7}{130}-\frac{9i}{130}$$
Also in your solution $7a-9b=1 \implies a= \dfrac{1+9b}{7} \neq \dfrac{9b}{7}$
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