复数z=l+cosθ+isinθ(π<θ<2π)的模为()。 A.A B.B C.C D.D 点击查看答案&解析手机看题 你可能感兴趣的试题 单项选择题 下列函数中,存在反函数的是()。 A.A B.B C.C D.D 点击查看答案&解析手机看题 单项选择题 用数学归纳法证明不等式的过程中,由“k”推导“k+1”时,不等式左边增加了...
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If z=costheta+isintheta is a root of the equation a0z^n+a2z^(n-2)++a(n-1)z^+an=0, then prove that a0+a1costheta+a2^cos2theta++ancosntheta=0 a1"sin"theta+a2^
确定极坐标方程 z=r(cos(theta)+isin(theta)) z=r(cos(θ)+isin(θ))z=r(cos(θ)+isin(θ)) 将方程重写为r(cos(θ)+isin(θ))=zr(cos(θ)+isin(θ))=z。 r(cos(θ)+isin(θ))=zr(cos(θ)+isin(θ))=z 将r(cos(θ)+isin(θ))=zr(cos(θ)+isin(θ))=z中的每一项除以cos(...
10.(复数的概念及充要条件)已知复数z=(cosθ-isinθ)(1+i),则”≮为纯虚数”的一个充分不必要条件是() A. \theta = \frac{ \pi }{4} \theta = \frac{ \pi }{2} C. \theta = \frac{3 \pi }{4} \theta = \frac{5 \pi }{4} ...
If Re ((z + 2i)/(z+4))= 0 then z lies on a circle with centre : 05:21 If z is a complex number satisfying the relation |z+ 1|=z+2(1+i), then... 03:24 The number of solutions of z^3+ z =0 is (a) 2 (b) 3 (c) 4 (d) 5 02:06 If z= costheta+isintheta...
Find the locus of center of circle of radius 2 units, if intercept cut on the x-axis is twice of intercept cut on the y-axis by the circle. View Solution Free Ncert Solutions English Medium NCERT Solutions NCERT Solutions for Class 12 English Medium ...
(c ) Z(P)=r(costheta+isintheta) Z(0)=sqrt(2|z|^(2))(cos(theta+(pi)/(4))+isin(theta+(pi)/(4))) =sqrt(2)r[cos(theta+(pi)/(4))+isin(theta+(pi)/(4))] From the figure, cos"(pi)/(4)=(2r^(2)+r^(2)-x^(2))/(2*sqrt(2r)*r)=(3r^(2)-x^(2))/(2...
Step 1: Substitute z into the equationSince z is a root, we can substitute z into the polynomial equation: a0(cosθ+isinθ)n+a1(cosθ+isinθ)n−1+a2(cosθ+isinθ)n−2+…+an−1(cosθ+isinθ)+an=0. Step 2: Use De Moivre's TheoremUsing De Moivre's Theorem, we can expre...
If z=1-costheta+isintheta,\ t h e n\ |z| is equal to a. 2sin(theta/2)... 02:13 If x+i y=(1+i)(1+2i)(1+3i),\ t h e n\ x^2+y^2 is: (a). 0 (b). 1 (c). 1... 03:17 If z=1/(1-cos theta-i sin theta), then Re(z) is a. 0 b. 1/2 c. c...