Complex Numbers help

Let z= 3 + ik and w= k + 7i, where k is a real number.

a.) express z/w in the form a+bi, where a and b are real numbers.

b.) for what value of k is z/w a real number?

thank you! any help would be greatly appreciated!

z/w= (3+ik)/(k+7i)
=(3+ik)(k-7i)/(k+7i)(k-7i)
=(3k-21i+ik²+7k)/(k²+49)
=[10k/(k²+49)] + i[(k²-21)/(k²+49)]
=......a..........................b

For z/w to be real, (k²-21)/(k²+49)=0, so k=±√21

z/w= (3+ik)/ ( k+7i)

= ((3+ik)/ ( k+7i) ) * (k-7i)/(k-7i)

= (3k+(-21+k^2)i +7k) / (k^2+49)

(z/w) = (10k +i(-21+k^2) ) / (k^2+49)

= 10k / (k^2+49) + i (-21+k^2) /(k^2+49)

b) A real number only , (-21+ k^2) /(k^2+49) =0
k^2=21
k= +- sqrt21

The real number must be N= +-10sqrt21/ 70
N= +-(1/7) sqrt21

z/w = (3+i*k)/(k + 7*i)

Multiply top and bottom by k - 7*i

z/w = (3+i*k)*(k - 7*i)/(k^2 + 7^2)

= (3*k + 7*k + i*k^2 - 21*i)/(k^2 + 7^2)

= (10*k + i*(k^2 - 21))/(k^2 + 49) <<< (a)

(b) This is real for (k^2 -21) = 0

k^2 = 21

ie k = +-sqrt(21) <<<

we konw complement of w = k - 7i

z/w = zw'/ww' = (3 + ik)(k-7i)/(k^2 + 7^2)

= ( 3k + 7k + (k^2- 21)i) /k^2 + 7^2))
= (10k + (k^2-21)i)/(k^2+ 49)

it is real when imaginary part is zero or k^2 = 21 or k = +/- sqrt(21)

z=3+i k , w = k+7i

(3+ik) / (k+7i) =(3+ik )(k-7i)/(k+7i)(k-7i) = [3k- 21i+ ik^2+7k](k^2+49)= [10k+(k^2-21)i]/(k^2+49)

z/w is real when k^2-21=0
k = +- Sqrt[21]