z=a+bi|z|=1+3i-z|z|=√(a^2+b^2)√(a^2+b^2)=1+3i-a-bi3-b=0b=3√(a^2+9)=1-aa^2+9=a^2-2a+1a=-4z=-4+3i(1+i)^2(3+4i)^2/z=(2i)(-7+12i)/(-4+3i)=(-14i-24)(-4-3i)/(16+9)=(54+128i)/25