Hmm thats the exact same question i had in my exam O_OI know this sound so simple, but I've never had to find the roots (apart for square) for something in cartesian form. I've always done it either just as a real number of just as an imaginary number.
Find the five-fifth roots of
and I'm tryna finish all my complex work tonight and I've got like 5 questions out of 153 I can't do
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Complex (1 Viewer)
- Thread starter nrlwinner
- Start date
ninetypercent
ninety ninety ninety
^ and did you get it right fullonoob?
I assume the denominator is 1-ik
Let ABC be the triangle with A,B,C being complex numbers u,v and that fraction (sorry cbf typing it)
To show that it is right angled, try showing vector AC=ki*(vector BC)
ie. try u-(the fraction) and v-(the fraction), simplify and see what you get
Let ABC be the triangle with A,B,C being complex numbers u,v and that fraction (sorry cbf typing it)
To show that it is right angled, try showing vector AC=ki*(vector BC)
ie. try u-(the fraction) and v-(the fraction), simplify and see what you get