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

1 answer:

7 0

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 $b=3i$ :

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

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Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b

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Step-by-step explanation:

From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .

So; as each place is fitted with either 0 or 1

$\mathbf{|R|= 2^{10}}$

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

$\mathbf{P(E) = \dfrac{|E|}{|R|}}$