Question #93684

U(x,y)+iv(x,y) a. Z^2+2iz b. e^z2 c. In(1+z)

Expert's answer

a)

z2=(x+iy)2=x2−y2+2ixyz^2=(x+iy)^2=x^2-y^2+2ixy

2iz=2i(x+iy)=−2y+2ix2iz=2i(x+iy)=-2y+2ix

z2+2iz=x2−y2−2y+2ixy+2ix−(x2−y2−2y)+i(2x+2xy)z^2+2iz=x^2-y^2-2y+2ixy+2ix-(x^2-y^2-2y)+i(2x+2xy)

U=x2−y2−2y;V=2x+2xyU=x^2-y^2-2y;\quad V=2x+2xy

b)

ez2=e(x+iy)2=ex2−y2+2ixy=ex2−y2e2ixy=ex2−y2(cos⁡(2xy)+isin⁡(2xy))e^{z^2}=e^{(x+iy)^2}=e^{x^2-y^2+2ixy}=e^{x^2-y^2}e^{2ixy}=e^{x^2-y^2}(\cos(2xy)+i\sin(2xy))

U=ex2−y2cos⁡(2xy);V=ex2−y2sin⁡(2xy)U=e^{x^2-y^2}\cos(2xy);\quad V=e^{x^2-y^2}\sin(2xy)

c)

ln(1+z)=ln(1+x+iy)ln(1+z)=ln(1+x+iy)

lnz=ln∣z∣+i(arg⁡(z)+2πk)lnz=ln|z|+i(\arg(z)+2\pi k)

∣z+1∣=(x+1)2+y2|z+1|=\sqrt{(x+1)^2+y^2}

arg⁡(z+1)=arctan⁡yx+1\arg(z+1)=\arctan\frac{y}{x+1}

ln(z+1)=ln(x+1)2+y2+i(arctan⁡yx+1+2πk)ln(z+1)=ln\sqrt{(x+1)^2+y^2}+i(\arctan\frac{y}{x+1}+2\pi k)

U=ln(x+1)2+y2;V=arctan⁡yx+1+2πkU=ln\sqrt{(x+1)^2+y^2} ;\quad V=\arctan\frac{y}{x+1}+2\pi k


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