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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

 Dec 12, 2017
 #1
avatar+98196 
+1

(a + bi)  (a - bi)   = 89

 

a^2   + b^2   =  89

 

One possibility

 

a  = 8,   b  = 5

 

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

 

 

cool cool cool

 Dec 12, 2017
 #2
avatar+21869 
+2

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,

 

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

 

laugh

 Dec 12, 2017
edited by heureka  Dec 12, 2017

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