已知非零复数z的主值为π/4,则z+1-i辐角主值范围.求详解,
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![已知非零复数z的主值为π/4,则z+1-i辐角主值范围.求详解,](/uploads/image/z/11641929-33-9.jpg?t=%E5%B7%B2%E7%9F%A5%E9%9D%9E%E9%9B%B6%E5%A4%8D%E6%95%B0z%E7%9A%84%E4%B8%BB%E5%80%BC%E4%B8%BA%CF%80%2F4%2C%E5%88%99z%2B1-i%E8%BE%90%E8%A7%92%E4%B8%BB%E5%80%BC%E8%8C%83%E5%9B%B4.%E6%B1%82%E8%AF%A6%E8%A7%A3%2C)
已知非零复数z的主值为π/4,则z+1-i辐角主值范围.求详解,
已知非零复数z的主值为π/4,则z+1-i辐角主值范围.求详解,
已知非零复数z的主值为π/4,则z+1-i辐角主值范围.求详解,
设z=r(cos∏/4+sin∏/4)
则z+1+i=(根2/2)r+1+((根2/2)r-1)i
则tana=((根2/2)r-1)/((根2/2)r+1)=1+2/(根2/2)r+1)
∵ r>0
∴ tana∈(-1,1)
∵a∈[0,2π)
∴a∈[0,π/4)∪(π/2,3π/4).
令 z=r*(cos(pi/4)+i sin(pi/4) );
y=z+1-i= r*sqrt(2)/2 (1+i) +1-i=r*sqrt(2)/2+1 +i (r*sqrt(2)/2-1);
令y=a +i b;
a= r*sqrt(2)/2+1 ;
b= (r*sqrt(2)/2-1);
ry=|y|= sqrt(a*a + b*b)=sqrt(r*...
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令 z=r*(cos(pi/4)+i sin(pi/4) );
y=z+1-i= r*sqrt(2)/2 (1+i) +1-i=r*sqrt(2)/2+1 +i (r*sqrt(2)/2-1);
令y=a +i b;
a= r*sqrt(2)/2+1 ;
b= (r*sqrt(2)/2-1);
ry=|y|= sqrt(a*a + b*b)=sqrt(r*r+2) ;
cos(arg(y)) =a/ry;
sin(arg(y)) =b/ry;
r->0 sin(arg(y)) =b/ry =-1/2;
r->∞sin(arg(y)) =b/ry =sqrt(2)/2;
sin(arg(y)) =b/ry=(r*sqrt(2)/2-1)/sqrt(r*r+2)在 (0,+∞) 之间是单调递减的;
所以 -1/2< b/ry
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