x = 1+2i = √5e^(i*arctan(2)) x³ = 5^(3/2) * e^(3i*arctan(2)) certainly not as clean
@itsphoenixingtime
3 ай бұрын
You can expand the exponential form out and then simplify. Cos(arctan x) and sin(arctan x) should give you some closed form expressions that can then be evaluated further
Пікірлер: 13
Somehow weird that a « looking positive » thing turns to be negative to the power of 3
@edwardpacman7082
2 ай бұрын
Think about (1+2i)*i and (1+2i)*i*i. Rotation is happening each time you mulply a complex number with none-zero imaginary part.
Or if you prefer... 5sqrt(5)e^(3i atan2)
this is pretty complex. Pun intended.
Sir please make a video on how to find intersection coordinates of two circles.
Can you evaluate (1+sqrt(2)i)^3
I think i am first viewer to watch you video
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Now solve it using radians.
x = 1+2i = √5e^(i*arctan(2)) x³ = 5^(3/2) * e^(3i*arctan(2)) certainly not as clean
@itsphoenixingtime
3 ай бұрын
You can expand the exponential form out and then simplify. Cos(arctan x) and sin(arctan x) should give you some closed form expressions that can then be evaluated further
@lumina_
3 ай бұрын
@@itsphoenixingtimetrue
What?