complex numbers/trig question (1 Viewer)

[YBK]

Well, this was part of a longer question where you had to find the minimum and maximum values of arg z. I did it using simultaneous equations, and the answer was good, but it took longer than it should so here's where I'm stuck, I'm trying to do the question through trigonometry.

tanx = 1/2
z=2(cos2x + isin2x)
Now could anyone please make that into a + ib form.



btw, the original question was:
|z-(2+i)| <= 1

P is the point in the region representing the complex number z0, with a maximum value of argz. Find z0 in the form a + ib.

Thanks!

 

[Riviet]

Draw a right triangle and label the two shorter sides 1 and 2, and label x as angle opposite the side length 1, such that tanx=1/2. Therefore hypotenuse is sqrt5.
So sinx=1/sqrt5
cosx=2/sqrt5

2(cos2x+isin2x)=2(cos²x-sin²x+2isinxcosx)
=2(4/5-1/5+2i.2/5)
=2(3/5+i.4/5)
=6/5+8i/5

 

[YBK]

Draw a right triangle and label the two shorter sides 1 and 2, and label x as angle opposite the side length 1, such that tanx=1/2. Therefore hypotenuse is sqrt5.
So sinx=1/sqrt5
cosx=2/sqrt5

2(cos2x+isin2x)=2(cos²x-sin²x+2isinxcosx)
=2(4/5-1/5+2i.2/5)
=2(3/5+i.4/5)
=6/5+8i/5

nice! thanks

That's easier than doing it by simultaneous equations