# Math Help - complex quadratic polynomial

QUESTION: Find a (complex) quadratic polynomial P(z)=a+bz+cz2 which interpolates the data: (-1,i),(0,i), (6 i, 0)

My Work:
So I have three different equations, using my x coordinate for my z values and my y coordinate as my P(z) values.
Equations:
i = a - b + c
i = a
0 = a + 6ib

I (tried) to solve each value by matrix reduction
__1_-1__1__i
__1_0__0 __i
__1_6i_-36_0

__1__-1___ 1____i
__0__ 1__(-6/i) _(-i/6)
__0__-1____1___0

(i continue to solve it until i get...):

a=i
b+(6i)c = -1/6
(1+6i)c = -1/6

if this IS right, i don't know how to actually solve the system
i end up getting b=c,
but if they answer has to be in x+iy how do i put it in this form??

thanks for the help!

2. Hello, williamb!

Find a complex polynomial $P(z)\:=\:a+bz+cz^2$ which has the points: $(-1,i),\;(0,i),\;(6 i, 0)$

Write your answers $a ,b, c$ in the form $x + iy$
From the given points, we have:

. . $\begin{array}{ccccccccc}P(\text{-}1) \:=\:i: & a + b(\text{-}1) + c(\text{-}1)^2 &=& i & \Rightarrow & a - b + c &=& i & {\color{blue}[1]} \\
P(0) \:=\:i: & a + b(0) + b(0^2) &=& i & \Rightarrow & a &=& i & {\color{blue}[2]}\\

P(6i) \:=\:0: & a + b(6i) + c(6i)^2 & = & 0 & \Rightarrow & a + 6ib - 36c &=& 0 & {\color{blue}[3]} \end{array}$

From [2]: . $\begin{array}{ccccccccc}{\color{blue}[1]}\text{ becomes:}& i - b + c &=& i & \Rightarrow & -b + c &=& 0 & {\color{blue}[4]} \\

{\color{blue}[3]}\text{ becomes:} & i + 6ib - 36c &=& 0 & \Rightarrow & 6ib - 36c &=& \text{-}i & {\color{blue}[5]}\end{array}$

$\begin{array}{cccc}\text{Multiply {\color{blue}[4]} by 36:} & \text{-}36b + 36c &=& 0 \\ \text{Add {\color{blue}[5]}:} & 6ib - 36c &=& \text{-}i \end{array}$

And we have: . $(\text{-}36+6i)b \:=\:\text{-}i \quad\Rightarrow\quad b \:=\:\frac{i}{36-6i} \;=\;\frac{\text{-}1+6i}{222} \;=\;\text{-}\tfrac{1}{222} + \tfrac{1}{37}i$

From [4], we have: . $c \:=\:b \quad\Rightarrow\quad c \:=\:\text{-}\tfrac{1}{222}+\tfrac{1}{37}i$

Therefore, the function is: . $P(x) \;=\;i + \left(-\tfrac{1}{222} + \tfrac{1}{37}i\right)x + \left(-\tfrac{1}{222} + \tfrac{1}{37}\right)x^2$