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# Find a complex number $z$ such that the real part and imaginary part of $z$ are both integers, and such that $$z\overline z = 89.$$

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Find a complex number $z$ such that the real part and imaginary part of $z$ are both integers, and such that $$$$z\overline z = 89$$.$$

the z with a line over it would equal to a-bi when z=a+bi

michaelcai  Dec 12, 2017
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#1
+85598
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(a + bi)  (a - bi)   = 89

a^2   + b^2   =  89

One possibility

a  = 8,   b  = 5

So...... z =  8 +  5i    is one solution

CPhill  Dec 12, 2017
#2
+19204
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Find a complex number $z$ such that the real part and imaginary part of $z$ are both integers, and such that $$z\overline z = 89.$$ The z with a line over it would equal to a-bi when z=a+bi

$$\begin{array}{|rcl|rcl|} \hline z\overline z &=& 89 & z\overline z &=& (a+bi)(a-bi)\\ && & &=& a^2+b^2 \\ a^2+b^2 &=& 89 \\ \hline \end{array}$$

The factorisation of $$89$$ is $$89^1$$, because $$89$$ is a prime number.

Because $$89 \equiv 1 \pmod 4$$ there are $$4\cdot ( \underbrace{1}_{\text{exponent of } 89} + 1 ) = 8$$ solutions,

$$\begin{array}{|r|r|r|r|r|} \hline & & a & b & z \\ \hline 1 & 8^2 + 5^2 = 89 & 8 & 5 & 8+5i \\ 2 & 5^2 + 8^2 = 89 & 5 & 8 & 5+8i \\ 3 & (-8)^2 + 5^2 = 89 & -8 & 5 & -8+5i \\ 4 & 5^2 + (-8)^2 = 89 & 5 & -8 & 5-8i \\ 5 & 8^2 + (-5)^2 = 89 & 8 & -5 & 8-5i \\ 6 & (-5)^2 + 8^2 = 89 & -5 & 8 & -5+8i \\ 7 & (-8)^2 + (-5)^2 = 89 & -8 & -5 & -8-5i \\ 8 & (-5)^2 + (-8)^2 = 89 & -5 & -8 & -5-8i \\ \hline \end{array}$$

heureka  Dec 12, 2017
edited by heureka  Dec 12, 2017

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