Browse · MATH Print → jmc algebra intermediate Problem Simplify 3−5i3+5i+3+5i3−5i. Solution — click to reveal We have that 3−5i3+5i+3+5i3−5i=(3−5i)(3+5i)(3+5i)(3+5i)+(3+5i)(3−5i)(3−5i)(3−5i)=9−25i29+15i+15i+25i2+9−25i29−15i−15i+25i2=9+259+30i−25+9−30i−25=34−32=−1716. Final answer -\frac{16}{17} ← Previous problem Next problem →