Find the inverse of $7+3i$ in $\mathbb{C}^*$.
I do not have a problem similar to this, therefore any help is appreciated.
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2What did you try? – Parcly Taxel Sep 26 '17 at 03:22
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1i can't find anything in my notes to help me even find something to try – rover2 Sep 26 '17 at 03:23
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1see correction, sorry – rover2 Sep 26 '17 at 03:27
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1With $\mathbb{C}^{+}$ you mean non-zero complex numbers, right? Then $\mathbb{C}^{*}$ is more common notation. – Krish Sep 26 '17 at 03:29
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yes you are right – rover2 Sep 26 '17 at 03:38
2 Answers
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HINT: If we try the conjugate (as our guess for the inverse) we have $$(7+3i)(7-3i)=49+9=58,$$ but we would really like this product to be $1$. Can we fix this?
Laars Helenius
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i understand the conjugate part, unsure where to go from where you ended – rover2 Sep 26 '17 at 04:04
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Divide by 58. The inverse of $7+3i$ should be $\frac{1}{58}(7-3i)$. – Laars Helenius Sep 28 '17 at 01:47
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An element of $C^{*}$ is of the form $a + bi$, where we cannot have $a=b=0$.
$a+bi$ is the inverse of $7+3i$ if $(a+bi)(7+3i) = 1$.
Expanding this gives us: $$7a + 7bi + 3ai + 3bi^{2} = 7a - 3b + (3a+7b)i= 1$$
Notice that 1 is purely real, so we must have $3a+7b = 0$. At the same time, we must have $7a-3b = 1$. Now, all that is left is to solve this pair of equations for appropriate integers $a$ and $b$.
eg123
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