Question

Complex Numbers

Answer 1

Solve the following quadratic equations:
(a) \[z ^{2}+1=0\]
(b) \[z ^{2}-4z+5=0\]

Answer 2

for (a)

Hint: \(\sf\Large a^2 + b^2 = (a+bi)(a-bi)\)

Answer 3

For the second one, it would be easier to use the quadratic formula.

Answer 4

This is what I got:
(a)\[z(z+1)=0\]
(b) \[(z-5)(z+1)\]

Answer 5

following your hint i got:
\[(z+i)(z-i)\]
how can i go on to find all possible roots of the equation from there?

Answer 6

If we had (x+1)(x+2) = 0

then we would have
x + 1 = 0, x = -1
x + 2 = 0, x = -2

In our case we have:

(z + i)(z - i) = 0

Therefore:

z + i = 0
z - i = 0

Solve for 'z'

Answer 7

also (b) is wrong :P

Answer 8

for (b) we should have gotten some complex roots

Answer 9

for (a) in polar form I got:
\[z _{1}=\cos \frac{ 3\pi }{ 2 }+isin \frac{ 3\pi }{ 2 }\]
and

Answer 10

\[z _{2}=\cos \frac{ \pi }{ 2 }+isin \frac{ \pi }{ 2 }\]

Answer 11

@TheSmartOne ?

Answer 12

>.<

Answer 13

Can you please help me out with (b)?

Answer 14

sorry, either I have not learned about polar form, or I totally forgot about it :p

Answer 15

np

Answer 16

@mathmale can you help?

Answer 17

\[z^2=-1\]
\[z^2=\iota^2,z=\pm \iota \]

Answer 18

Okay, so what about (b)?

Answer 19

use the solution for x= from the standard quadratic ax^2+bx+c

you remember the quadratic formula x=[-b +- root(b^2 - 4*a*c)] / (2a)

Answer 20

yes I know that formula

Answer 21

that will tell you x values when the function is zero, x intercepts

that will help you form a factored form of that thing,

Answer 22

z^2 - 4z + 5,

Answer 23

@surjithayer I got that the first time and thesmatone said it was wrong.

Answer 24

the root will be of a negative, complex solutions

Answer 25

ok i'm giving it a go

Answer 26

\[z=\frac{ 4\pm \sqrt{(-4)^2-4*1*5} }{ 2*1 }=\frac{ 4\pm \sqrt{-4} }{ 2 }=\frac{ 4\pm2\iota }{ 2 }=2\pm \iota \]

Answer 27

@DanJS I got 2+i and 2-i. However this was off a calculater. is there any way to do it without?

Answer 28

yeah just shown ^

Answer 29

great @surjithayer you answered it already!

Answer 30

works for any real or complex solution, depending what that b^2-4*a*c tells ya

Answer 31

thanks for your help.

Answer 32

ah, we didn't even need the polar form :P

Answer 33

hah, i saw that and wondered where it was goin

Answer 34

same x'D