Question

How to find the roots to a equation with complex coefficients? I tried one, but i'm stuck

Answer 1

\[2x^2-ix-2+i\]

Answer 2

You can use the quadratic formula with a=2, b= -i and c= (-2+i)
But you will have to take the square root of a complex number. See
https://en.wikipedia.org/wiki/Square_root#Algebraic_formula
for one way to do this.

Answer 3

I did it..but I got \[x^4+17x^2-16\]

Answer 4

and idk how to solve that

Answer 5

my prof does it differently

Answer 6

and i dont get his way

Answer 7

@phi

Answer 8

\[x=\frac{ \iota \pm \sqrt{\left( -\iota \right)^{2}-4*2\left( -2+\iota \right)} }{ 2*2 }\]
\[=\frac{ \iota \pm \sqrt{-1+16-8\iota } }{ 4 }\]
\[=\frac{ \iota \pm \sqrt{15-8\iota} }{4 }=\frac{ \iota \pm \sqrt{16-1-8\iota} }{ 4 }\]
\[=\frac{ \iota \pm \sqrt{\left( 4 \right)^{2}+\iota ^{2}-2*4*\iota} }{4 }\]
\[=\frac{ \iota \pm \left( 4-\iota \right) }{4 }\]
now you can solve to find the two values.

Answer 9

omg..nvm..i switched the -ive in the -4 part of the quadratic formula to a +

Answer 10

@surjithayer

Answer 11

can u help me with something

Answer 12

I'm getting the wrong roots

Answer 13

can someone check what I'm doing wrong

Answer 14

hint : use a=x^2

Answer 15

then you will have
a^2+17a-16=0
(a+1)(a-16)=0
a=-1 ,a=16
x^2=-1 ---> x=+-i
x^2=16 --->x=+-4

Answer 16

or
from the first

it seems x^2=-1 is a factor
check it
or see below
x^2=-1
x^4+17x^2-16
=(-1)^2+17(-1)-16=0
so you can divide x^4+17x^2-16 by x^2+1

Answer 17

2x^2−ix−2+i
=2x^2-2−ix+i
=2(x^2-1) -i(x-1)
now you can factor (x-1)

Answer 18

ok. when i put it into the quadratic form and try to find the roots. Do i use all 4 of the roots that i get

Answer 19

could i post my work and you check it over?

Answer 20

\[2x^2-ix-2+i =0 \rightarrow \Delta=\sqrt{i^2-4(2)(-2+i)}\]
\[\Delta=\sqrt{i^2-4(2)(-2+i)} =\sqrt{-1-8(-2+i)}=\sqrt{-1+16-8i}\]
\[\sqrt{-1+16-8i}=\sqrt{(4-i)^2}=4-i\]

Answer 21

i posted my work

Answer 22

\[x =\frac{ +i \pm \Delta }{ 4 }=\frac{ +i \pm(4-i) }{ 4 }\]

Answer 23

ok..i looked up a video and it said that i only use 1 of the sets of roots

Answer 24

could u look at my work and correct me where im going wrong

Answer 25

it was very unclear
but i find your mistake
(-1+16-8i)=(4-i)^2 not (-1+4i)^2

Answer 26

(4-i)^2=16+i^2-8i

but
(-1+4i)^2=1+16i^2-8i=1-16=8i

Answer 27

how did u get (4-i)

Answer 28

-1+16 -8i =i^2 +4 ^2 -2 *4 * i

Answer 29

???

Answer 30

could u write it using the equation function on here

Answer 31

function !
it was (a-b)^2 =a^2 +b^2 -2 a b
just!

Answer 32

I get 15+8i...and then i evaluate it on the far right side of my work....

Answer 33

can u look there and see what im doing wrong

Answer 34

ok begin with 15+8i
\[15+8i =\sqrt{15^2+8^2} e^{i \theta}\]
\[\sqrt{225+64} e^{i \theta}=17 e^{i \theta} \]
\[\theta = \tan ^{-1} \frac {y}{x} =\tan ^{-1} \frac {8}{15} \]

Answer 35

can u not use e..i didnt learn it