Can anyone help me with this?

[wintry trail]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations

[main dust]

you're just converting 3+4i into modulus argument Form, |3+4i|=5 and the arg is arctan(4/3) then you just have 5e^(iarctan(4/3)) in the root, distribute the root and then revert back to a+bi form with re^ix=cosx+isinx

@timid turret

[timid turret]

But I don't get how they found arctan(4/3) and idk how to revert it to a+bi

[main dust]

you find arg=arctan(4/3) by constructing a right angle triangle of side lengths 3,4,5

tanx=o/a so arctan(o/a)=x

hence arctan(4/3)=arg

[main dust]

you revert by subbing in your end modulus (√5 after distributing the root) and your x which is your arctan(4/3)/2

[storm hamlet]

PDF File (apex predator)

[main dust]

where r is your modulus and x is your argument

@timid turret

[timid turret]

But it only gives me for sqrt5cos(2/3)= 2.2359... and for sqrt5sin(2/3)=0.02601...
So if I round it it will give me (2+0)
Since the second part is closer to 0 ?

[main dust]

it's not 2/3 what

you can distribute the half to the inside

you have to do 0.5arctan(4/3)

@timid turret

[timid turret]

But still try it on calculator it doesn't give the result at the end

[main dust]

it did for me

√5cos(0.5arctan(4/3))+i√5sin(0.5arctan(4/3))=2+i