Let's split them up into two separate complex numbers to start with, one being the numerator, #2i+5#, and one the denominator, #-7i+7#.

We want to get them from linear (#x+iy#) form to trigonometric (#r(costheta + isintheta)# where #theta# is the argument and #r# is the modulus.

For #2i+5# we get

#r = sqrt(2^2 + 5^2) = sqrt29#

#tantheta = 2/5 -> theta = arctan(2/5) = 0.38 " rad"#

and for #-7i+7# we get

#r = sqrt((-7)^2+7^2) = 7sqrt2#

Working out the argument for the second one is more difficult, because it has to be between #-pi# and #pi#. We know that #-7i+7# must be in the fourth quadrant, so it will have a negative value from #-pi/2 < theta < 0#.

That means we can figure it out simply by

#-tan(theta) = 7/7 = 1 -> theta = arctan(-1) = -0.79 " rad"#

So now we've got the complex number overall of

#(2i+5)/(-7i+7) = (sqrt29(cos(0.38)+isin(0.38)))/(7sqrt2(cos(-0.79)+isin(-0.79)))#

We know that when we have trigonometric forms, we divide the moduli and subtract the arguments, so we end up with

#z = (sqrt29/(7sqrt2))(cos(0.38+0.79)+isin(0.38+0.79))#

# = 0.54(cos(1.17)+isin(1.17))#