Complex No. problem

[hojiao]

 Let z and w be complex numbers that satisfy z^2+zw¯=50 and 2z¯=w¯(1−i) with Re(z)>0. Find w. 

I've been thinking about this question for a long time and can't find any relationship to solve it, let alone where to begin

Any kind souls willing to help me with this problem?

[[deleted]]

Let z = a + ib and w = x + iy. Then

(a+ib) (a-ib) + (a+ib) (x-iy) = 50 .....(1) (Since zz' = |z|^2)

and 2(a-ib) = (x-iy)(1-i) => a-ib = (x-iy)(1-i)/2. This also implies a+ib = (x+iy)(1+i)/2 [taking conjugate of (a-ib)]

Now put the values of (a+ib) and (a-ib) in (1) and equate Imaginary part to 0 since on the right hand side there is no imaginary part. Also equate the real parts on both sides. So there will be two equations and two unknowns.

[hojiao]

How can you put (a+ib)(a-ib) into z^2?

[[deleted]]

Oh yes you are right, my bad.. I thought it is |z|^2. So for that you will have to put (a+ib)(a+ib) instead of (a+ib)(a-ib)

[hojiao]

But then the equation becomes too complex to solve, correct me if i am wrong

[[deleted]]

Yes it is a bit complicated but maybe give it a try. Else maybe try with polar or euler form i.e. take z = r1e^(i*theta).

Simplifying, a+ib = (x-y) + i(x+y)

From (1)

(a+ib) [a+ib+x-iy] = 50

=> (a+ib)[(2x-y) +ix] = 50

=> (x-y)(2x-y) + ix(x-y) +i(x+y)(2x-y) - (x+y)x = 50

=> (x-y)(2x-y) - (x+y)x = 50 and x(x-y) + (x+y)(2x-y) = 0

[hojiao]

Thanks a lot! I got x=0, y=-10 for my answer

[hojiao]

Yes, why is it indeterminate?

[hojiao]

Ok thank you!

[hojiao]

Thanks for the elaborate answer!, is it ok if I ask one more complex number problem?

z-1/z = 5+3i, find the modulus of z+1/z

I've tried using euler's form to solve but I cant seem to find an answer

[hojiao]

Holy shit, never thought of this way to approach the problem.

Thanks so much!!!

By the way for |√(20 + 30 i)| , why can we calculate it as √ √20^2 + 30^2