How to solve complex equation with i in the bx term?

Question

First of all, sorry for the English.

I have never asked question about Maths in English before, so I don't know if my problem is understandable.

So, basically I don't know anything about this stuff, I have never learnt stuff like this before and only in university did they spill the tea to me, so I desperately need someone to help me understand these equations.

For example:

a) $z^3-(3+2i)z^2+(1+3i)z=0$

b) $ -iz^3 = 64 $

c) $z^4+2(i+1)z^2+i=0$ (I know that i could make a new variable eg. $y=z^2$)

Should I use trigonometric format or algebraic format? Where do I even start?

I have not learn any formula specifically to the $x^3$ equation.

EDIT:

I cant use a calculator either, nor any help from anything else.

For b), just multiply by $-i$, so we solve $z^3=-64i$. I suppose you know the third roots for complex numbers. Start with the roots of unity $1,\omega,\omega^2$ with $\omega=e^{2\pi i/3}$. For the cubic equation, also over $\Bbb C$, see wikipedia. — Dietrich Burde, Sep 19 '23 at 14:32
And c) is just a quadratic equation in $y=z^2$. You can use the usual formula. — Dietrich Burde, Sep 19 '23 at 14:36
But then i have to calculate the $sqrt(80-18i)$ and i cant do that. — GooDinosaur, Sep 19 '23 at 14:39
And a) si just $zP(z)=0,$ where $P(z)$ is a quadratic polynomial. I.e. $z=0$ or $P(z)=0.$ — Anne Bauval, Sep 19 '23 at 14:44
a) z=0 is one solution. Use the quadratic formula for the others. b) $i^3=-i$. Take the cube root of both sides. c) substitute $u=z^2$ then use the quadratic formula to find u. Then take the square root of those solutions.
$\sqrt{a+bi} = \sqrt{\sqrt{a^2+b^2}} e^{i(\arctan b/a)/2}$ — TurlocTheRed, Sep 19 '23 at 14:53

score 0 · Answer 1 · edited Sep 19 '23 at 16:20

Thank you everyone! I could solve some of it or at least it gave me a place to start.

The progress I made:

Problem a)

I know that $z^3−(3+2i)z^2+(1+3i)z=0$ only can be zero if:

A) $z=0$

B) $-(3+2i)z^2=0$ $\rightarrow$ $ -3z-2iz=0$ ?

C) $(1+3i)z=0$ $\rightarrow$ $z+3iz=0$ ?

I'm stuck at this stage, cause it's really weird that everything is 0.

Problem b) [solved]

A) $2+2\sqrt3\cdot i$

B) $2-2\sqrt3\cdot i$

C) $-4$

Problem c)

$x_{1,2}=\frac{-(9+i)\pm\sqrt{80-18i}}{2}=\frac{-9-i\pm9-i}{2}\rightarrow x_1=-i;x_2-9$

But I'm not sure in this answer and I could not calculate the $\sqrt{80-18i}$ in trigonometric form.

How to solve complex equation with i in the bx term?

1 Answers1

Problem a)

Problem b) [solved]

Problem c)