Prove that 1+i is a root of the equation z^4+3z^2-6z+10=0

Question

apoorvk · Answer

substitute (1+i) in place of 'z'. then simplify. you ll see it simplifies to 0.

anonymous · Answer

I can't see that...?

mani_jha · Answer

Convert 1+i into the polar form and then substitute.

$$\theta=45$$
$$1+i=\cos45+isin45$$
Use this identity:
$$(\cos \theta+i \sin \theta)^{n}=\cos(n \theta)+isin(n \theta)$$
Get it?

apoorvk · Answer

z is any no. complex which is a root of the given expression. now root, as you may remember, means a no. which when substituted in the variable of any expression reduces the expression to zero. so here put in z=1+i

so you get:

$$(1+i)^4 + 3(1+i)^2 - 6(1+i) +10 = 0$$

now see after simplifying whether both the real and imaginary parts add up to zero.

you need to solve each term of the equation

anonymous · Answer

I haven't learned polar form yet..

anonymous · Answer

I understand apporvk :)

apoorvk · Answer

now i think using polar form when you have learnt it will actually make it pretty easy. right mani!!

anonymous · Answer

synthetic division works also