Question

Write an equation the has the given solutions- 1+3i, 1-3i Show work please!

Answer 1

We know that z₁=1+3i and z₂=1-3i are solution, so:

$(z-z_1)(z-z_2)=0\\\\\big(z-(1+3i)\big)\big(z-(1-3i)\big)=0\\\\(z-1-3i)(z-1+3i)=0\\\\ \big((z-1)-3i\big)\big((z-1)+3i\big)=0$

Now we use $(a-b)(a+b)=a^2-b^2$ where $a=z-1$ and $b=3i$:

$\big((z-1)-3i\big)\big((z-1)+3i\big)=0\\\\ (z-1)^2-(3i)^2=0\\\\z^2-2z+1-9i^2=0\qquad [i^2=-1]\\\\ z^2-2z+1+9=0\\\\\boxed{z^2-2z+10=0}$