Question

How do you solve this complex number equation?

Answer 1

\[z ^{4}+(1-2i)z ^{2}-2i=0\]

Answer 2

like a tipical quadratic equation, using discriminant

Answer 3

z^4 + (1 - 2i)z^2 - 2i = 0
z^4+ z^2 - 2iz^2 - 2i = 0
z^2(z^2+ 1) - 2i(z^2 + 1) = 0
(z^2-2i)(z^2+ 1) = 0

z^2-2i =0 --> z^2=2i --> z=√(2i)
2i=(1+i)(1+i) hence z=√(2i)=±(1+i)

or
(z^2+ 1) = 0
(z-i)(z+i)=0
z=±i

(sorry for late reply)

Answer 4

I was busy watching a movie

Answer 5

Thank you very much!

Answer 6

Not a problem:)

Answer 7

I need to ask something else

Answer 8

Yes sure:) I am not very good at math, but I am creative ;)

Answer 9

Why did you write only z=√(2i) and not both +√(2i) and - √(2i)

Answer 10

?

Answer 11

Well, I meant that, I was jus typing in a little rush. I wanted to say z=±√2i

By the way,

Hold ALT, click 2 5 1 (using numbers on the right of your keyboard) and release ALT. You get √

Hold ALT, click 7 5 3 (using numbers on the right of your keyboard) and release ALT. You get ±

Answer 12

ok, and I think you made a mistake in the last part...

Answer 13

(z^2+ 1) = 0
(z-i)(z+i)=0
It's wrong because you can do it only when (z^2- 1) and not the other way round

Answer 14

Yes, I did. Sorry

Answer 15

No I didn't... sorry, you confused me a little.

Answer 16

2 common forms are the following (keep making errors, sorry -;(

\(\Large\color{red}{ \bf (a^2- b^2) =(a-b)(a+b) }\)

\(\Large\color{green}{ \bf (a^2+ b^2) =(a-ib)(a+ib) }\)
\(\Large\color{green}{ \bf (b^2+ a^2) =(b-ia)(b+ia) }\)

Answer 17

ah ok. thanks

Answer 18

Anytime !