First of all we have to convert these two numbers into trigonometric forms.
If (a+ib) is a complex number, u is its magnitude and alpha is its angle then (a+ib) in trigonometric form is written as u(cosalpha+isinalpha).
Magnitude of a complex number (a+ib) is given bysqrt(a^2+b^2) and its angle is given by tan^-1(b/a)
Let r be the magnitude of (2-3i) and theta be its angle.
Magnitude of (2-3i)=sqrt(2^2+(-3)^2)=sqrt(4+9)=sqrt13=r
Angle of (2-3i)=Tan^-1(-3/2)=theta
implies (2-3i)=r(Costheta+isintheta)
Let s be the magnitude of (-3-7i) and phi be its angle.
Magnitude of (-3-7i)=sqrt((-3)^2+(-7)^2)=sqrt(9+49)=sqrt58=s
Angle of (-3-7i)=Tan^-1((-7)/-3)=Tan^-1(7/3)=phi
implies (-3-7i)=s(Cosphi+isinphi)
Now,
(2-3i)(-3-7i)
=r(Costheta+isintheta)*s(Cosphi+isinphi)
=rs(costhetacosphi+isinthetacosphi+icosthetasinphi+i^2sinthetasinphi)
=rs(costhetacosphi-sinthetasinphi)+i(sinthetacosphi+costhetasinphi)
=rs(cos(theta+phi)+isin(theta+phi))
Here we have every thing present but if here directly substitute the values the word would be messy for find theta +phi so let's first find out theta+phi.
theta+phi=tan^-1(-3/2)+tan^-1(7/3)
We know that:
tan^-1(a)+tan^-1(b)=tan^-1((a+b)/(1-ab))
implies tan^-1(-3/2)+tan^-1(7/3)=tan^-1(((-3/2)+(7/3))/(1-(-3/2)(7/3)))
=tan^-1((-9+14)/(6+21))=tan^-1((5)/(27))
implies theta +phi=tan^-1((5)/(27))
rs(cos(theta+phi)+isin(theta+phi))
=sqrt13sqrt58(cos(tan^-1 (5/27))+isin(tan^-1 (5/27)))
=sqrt754(cos(tan^-1 (5/27))+isin(tan^-1 (5/27)))
This is your final answer.
You can also do it by another method.
By firstly multiplying the complex numbers and then changing it to trigonometric form, which is much easier than this.
(2-3i)(-3-7i)=-6-14i+9i+21i^2=-6-5i-21=-27-5i
Now change -27-5i in trigonometric form.
Magnitude of -27-5i=sqrt((-27)^2+(-5)^2)=sqrt(729+25)=sqrt754
Angle of -27-5i=tan^-1(-5/-27)=tan^-1(5/27)
implies -27-5i=sqrt754(cos(tan^-1(5/27))+isin(tan^-1(5/27)))
