(-3-4i)/(5+2i)=-(3+4i)/(5+2i)−3−4i5+2i=−3+4i5+2i
z=a+biz=a+bi can be written as z=r(costheta+isintheta)z=r(cosθ+isinθ), where
- r=sqrt(a^2+b^2)r=√a2+b2
- theta=tan^-1(b/a)θ=tan−1(ba)
For z_1=3+4iz1=3+4i:
r=sqrt(3^2+4^2)=5r=√32+42=5
theta=tan^-1(4/3)=~~0,927θ=tan−1(43)=≈0,927
For z_2=5+2iz2=5+2i:
r=sqrt(5^2+2^2)=sqrt29r=√52+22=√29
theta=tan^-1(2/5)=~~0.381θ=tan−1(25)=≈0.381
For z_1/z_2z1z2:
z_1/z_2=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))z1z2=r1r2(cos(θ1−θ2)+isin(θ1−θ2))
z_1/z_2=5/sqrt(29)(cos(0.921-0.381)+isin(0.921-0.381))z1z2=5√29(cos(0.921−0.381)+isin(0.921−0.381))
z_1/z_2=5/sqrt(29)(cos(0.540)+isin(0.540))=0.79+0.48iz1z2=5√29(cos(0.540)+isin(0.540))=0.79+0.48i
Proof:
-(3+4i)/(5+2i)*(5-2i)/(5-2i)=-(15+20i-6i+8)/(25+4)=(23+14i)/29=0.79+0.48i−3+4i5+2i⋅5−2i5−2i=−15+20i−6i+825+4=23+14i29=0.79+0.48i
