COmplex Numbers Mod-arg (1 Viewer)

[kevda1st]

Convert 6(cos(3pi/4) + iSin(3pi/4)) into cartesian form
ie 6(Cos135 + isin135)

 

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回复: COmplex Numbers Mod-arg

Convert 6(cos(3pi/4) + iSin(3pi/4)) into cartesian form
ie 6(Cos135 + isin135)

6e^iθ = 6[cos(3π/4) + isin(3π/4)]

Cartesian form is x + iy and x = rcosθ and y = rsinθ
.:
x = 6cos3π/4
x = 6 . -1/√2
x = -3√2

y = 6sin3π/4
y = 6 . 1/√2
y = 3√2

Cartesian form is -3√2 + i3√2 amirite???

 

[gurmies]

6(cos(3pi/4) + iSin(3pi/4))

= 6(-1/root2 + (1/root2)i)

= -6/root2 + (6/root2)i

= -(6root2)/2 + ((6root2)/2)i

= -3root2 + (3root2)i

 

[x.Exhaust.x]

Cartesian form: x+iy

6[(cos3pi/4) + isin(3pi/4)]

= 6[cos(135) + isin(135)]

= 6[cos(-45) + isin(45)] (remember ASTC)

= (6 x -1/root 2) + 6i(1/root 2)

= -6/root 2 + 6i/root 2 (rationalise)

= (-3root2) + (3root2)i

 

[tacogym27101990]

Re: 回复: COmplex Numbers Mod-arg

6e^iθ = 6[cos(3π/4) + isin(3π/4)]

Cartesian form is x + iy and x = rcosθ and y = rsinθ
.:
x = 6cos3π/4
x = 6 . -1/√2
x = -3√2

y = 6sin3π/4
y = 6 . 1/√2
y = 3√2

Cartesian form is -3√2 + i3√2 amirite???

i believe 6cis(3pi/4) equals 6e^i3pi/4

and it is so unnecessary to quote that rule

 

[kevda1st]

Cartesian form: x+iy

6[(cos3pi/4) + isin(3pi/4)]

= 6[cos(135) + isin(135)]

= 6[cos(-45) + isin(45)] (remember ASTC)

= (6 x -1/root 2) + 6i(1/root 2)

= -6/root 2 + 6i/root 2 (rationalise)

= (-3root2) + (3root2)i

ohhhhhh, forgot to rationalise....
Thanks for the help =D