Compute
$\frac{(1+2i)^{2011}}{(2-i)^{2011}}$

Solution:
$\frac{1+2i}{2-i}=i$

hence
$\frac{(1+2i)^{2011}}{(2-i)^{2011}}=i^2011=-i$

Find all possible values of
$1^{1+i}$

Solution:
$1^{1+i}=e^{2\pi i k (1+i)}=e^{2\pi i k - 2\pi k}=e^{2\pi i k}e^{-2\pi k}=e^{-2\pi k}$

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
$\sum\limits_{n=1}^{\infty} \frac{(\pi i)^n}{n!}$

Solution:
$\sum\limits_{n=1}^{\infty} \frac{(\pi i)^n}{n!}=\sum\limits_{n=0}^{\infty} \frac{(\pi i)^n}{n!} - 1=e^{\pi i} - 1 = -2$

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
$\lim_{z\to 0} \frac{f(z)-f(0)}{z-0}=\lim_{(x,y)\to (0,0)}\frac{x^2y+ixy^2}{x+iy}=\lim_{(x,y)\to (0,0)} xy=0$
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
$u_x+v_x=-1$
$u_y+v_y=1$
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
$\int \frac{dz}{z-i}$
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
$\int \operatorname{Re}z \,dz [[math] over the unit circle |z|=1. Solution: Green's theorem implies that$
\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\}
$under f(z)=cos z. Solution: for real values of z the value of cos z is a real number between -1 and 1. In particular it isn't 5. Show that w=5 is in the image of$
\{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