Thursday

July 24, 2014

July 24, 2014

Posted by **Kate** on Friday, May 17, 2013 at 9:11am.

a) -5 + 5i

b) -5

c)-3 -4i

- TRIGONOMETRY -
**Steve**, Friday, May 17, 2013 at 9:51amthink back to your polar coordinates. Same thing.

(a+bi) = r cisθ

where

tanθ = y/x

r^2 = a^2 + b^2

-5+5i

r^2 = 5^2+5^2, so r=5√2

tanθ = -5/5 = -1

Now, cosθ=x/r sinθ=y/r

since x<0 and y>0, θ is in II, so

θ = 3π/4

-5 = -5+0i = 5 cis π

-3-4i = 5 cis π+arctan(-4/-3)

because we are in QIII

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