Find |w| without finding w? w is a complex number that satisfy: 5w3−3i|w|2−2i=0 Can we find |w| without finding w or easy root?

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Find |w| without finding w?  w  is a complex number that satisfy:  5w3−3i|w|2−2i=0  Can we find |w| without finding w or easy root?

nick1337 · Accepted Answer

Step 1 Take w=reiϕ. Then you get: 5r3ei3ϕ−3ir2−2i=0 Which can be rearranged to: 5r3ei3ϕ=i(3r2+2) divide by i on both sides: 5r3ei3ϕ−iπ2=3r2+2⏟∈R since RHS is real and positive, LHS is real and positive, i.e. ei3ϕ−iπ2=1 Thus we get: 5r3−3r2−2=0 and we want the positive real roots to this equations, which is r=1 EDIT: Fixed an error.

Vasquez · Accepted Answer

Step 1 For any solution w, the two complex numbers 5w3 and (3|w|2+2)i are the same hence their modulus are the same: 5(w)3=3(w)2+2 This means that the modulus r of any solution w is such that 5r3−3r2−2=0 Here ends the answer to the OP question. Naturally one can go further since an obvious solution of the r equation is r=1. Ytnce 5r3−3r2−2=(r−1)(5r2+2r+2) and, since 5r2+2r+2=0 has no real solution, every solution of the w equation is such that |w|=1 All this does not say that there exists any solution w at all but one can go further and note that a complex number w such that |w|=1 is a solution iff the argument of w3 is π2 hence the argument of w is π6 or 5π6 or 3π2 Finally the three solutions are eiπ6=3+i2 ei5π6=−3+i2 and ei3π2=−i

alenahelenash · Accepted Answer

Since the second and third terms are pure imaginary, the angle of the solution is either π6, 5π6, or 3π2. To solve for the magnitudes, just solve the real equation 5w3−3w2−2=0 which clearly has a real positive root at 1.

Find |w| without finding w? w is a complex number

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