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

#1**+1 **

(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**+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,

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

heureka Dec 12, 2017