Write your answer in the form a+bi with real numbers a,b

Solution:

hence

Find all possible values of

Solution:

Compute the exact value of cos(i ln 2) - i sin (i ln 2).

Solution: cos(i ln 2) - i sin (i ln 2) = cos(-i ln 2)+ i sin(- i ln 2) = e^(i (-i ln 2))=e^(ln2)=2

Compute the exact value of

Solution:

Find all the points where the function f(x+iy)=x^2y+i xy^2 is complex-differentiable.

Cauchy-Riemann equations with u(x,y)=x^2 y, v(x,y)=xy^2 yield
2xy=2xy
and
x^2=-y^2

The second equation x^2+y^2=0 has only one solution, x=y=0. Hence f can have a complex derivative only at z=0.

At z=0 we compute

so the derivative does exist at z=0.

Suppose u(x,y),v(x,y) are real-valued functions such that the function f(x+iy)=u(x,y)+iv(x,y) is analytic on the whole complex plane. Suppose also that the equation u(x,y)+v(x,y)=y-x holds for all x,y and f(0)=1. Find f(2+i).

Differentiating u+v=y-x with respect to x and with respect to y we get

Together with Cauchy-Riemann equations u_x=v_y, u_y=-v_x these equations imply u_y=1, u_x=0, v_x=-1,v_y=0. Hence u(x,y)=y+a, v(x,y)=-x+b for some real a,b. Plugging this into f(0)=0 we find that a=b=0. Hence f(2+i)=1-2i.

Find the value of

over the straight line segment from z=1+i to z=0.

Solution: The integral is equal to the difference of ln(z-i) at the endpoints, where ln(z-i) is chosen so that it is continuous on the line segment. As z moves from 1+i to 0, the argument of z-i changes from 0 to -pi/2. The absolute value of z-i is the same at both endpoints. Hence the difference of ln(z-i) = ln|z-i|+i arg (z-i) at the endpoints is -pi i/2.

Find the value of the integral

\int \operatorname{Re}z \,dz=\int x \, (dx+i dy)= \int \int 0 dx dy + i \int \int 1 \, dx dy=i \mbox{Area(circle)}=\pi i
[[math]Prove that w=5 is not in the image of math\{z: \operatorname{Im} z=0\}

Write your answer in the form a+bi with real numbers a,b

Solution:

hence

Find all possible values of

Solution:

Compute the exact value of cos(i ln 2) - i sin (i ln 2).

Solution: cos(i ln 2) - i sin (i ln 2) = cos(-i ln 2)+ i sin(- i ln 2) = e^(i (-i ln 2))=e^(ln2)=2

Compute the exact value of

Solution:

Find all the points where the function f(x+iy)=x^2y+i xy^2 is complex-differentiable.

Cauchy-Riemann equations with u(x,y)=x^2 y, v(x,y)=xy^2 yield

2xy=2xy

and

x^2=-y^2

The second equation x^2+y^2=0 has only one solution, x=y=0. Hence f can have a complex derivative only at z=0.

At z=0 we compute

so the derivative does exist at z=0.

Suppose u(x,y),v(x,y) are real-valued functions such that the function f(x+iy)=u(x,y)+iv(x,y) is analytic on the whole complex plane. Suppose also that the equation u(x,y)+v(x,y)=y-x holds for all x,y and f(0)=1. Find f(2+i).

Differentiating u+v=y-x with respect to x and with respect to y we get

Together with Cauchy-Riemann equations u_x=v_y, u_y=-v_x these equations imply u_y=1, u_x=0, v_x=-1,v_y=0. Hence u(x,y)=y+a, v(x,y)=-x+b for some real a,b. Plugging this into f(0)=0 we find that a=b=0. Hence f(2+i)=1-2i.

Find the value of

over the straight line segment from z=1+i to z=0.

Solution: The integral is equal to the difference of ln(z-i) at the endpoints, where ln(z-i) is chosen so that it is continuous on the line segment. As z moves from 1+i to 0, the argument of z-i changes from 0 to -pi/2. The absolute value of z-i is the same at both endpoints. Hence the difference of ln(z-i) = ln|z-i|+i arg (z-i) at the endpoints is -pi i/2.

Find the value of the integral

\int \operatorname{Re}z \,dz=\int x \, (dx+i dy)= \int \int 0 dx dy + i \int \int 1 \, dx dy=i \mbox{Area(circle)}=\pi i

[[math]Prove that w=5 is not in the image of math\{z: \operatorname{Im} z=0\}

\{z: \operatorname{Im}z >0\}

math

under f(z)=e^z.

Solution: 5=e^(ln 5+ 2pi i) and Im(ln 5 +2 pi i)=2 pi > 0

math