Week 1 Worksheet - Solutions

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Complex Numbers

    1. |2i|=2. The arguments are π2+2nπ for any nZ. Arg(2i)=π2.
    2. |1i3|=(1)2+(3)2=2. The arguments are 4π3+2nπ for any nZ. The principal argument is 2π3.
    3. |4|=4. The arguments are π+2nπ for any nZ. Arg(4)=π.

Arithmetic

  1. Calculate using the rules of arithmetic. (2+3i)(5+i)=(2×(5)3×1)+i(2×1+3×(5))=1313i
    1. 5+24i
    2. 2+3i34i3+4i3+4i=6+17i25=625+1725i
    3. 15i3i13i13i1=16+2i10=85+15i
    4. 1i1+i1i1ii+2=2i2i+2=22i
    5. i (becuase ii=1)
    1. Write z=x+iy so that the equation becomes x2y2+2xyi=512i. Real and imaginary parts must be equal, so x2y2=5 and 2xyi=12i. By inspection we can take x=2,y=3 or x=2,y=3. The solutions are z=2+3i and z=23i.

    2. Completing the square gives (z+2)24=12+6i so (z+2)2=8+6i. Writing z+2=x+iy gives the equations x2y2=8 and 2xyi=6i. By inspection, we can take x=1,y=3 or x=1,y=3. The solutions are therefore 1+3i and 33i.

    1. Calclate Re(z+w)=Re((x+iy)+(u+iv))=Re((x+u)+i(y+v))=x+u=Re(z)+Re(w)
    2. Calculate Im(zw)=Re((x+iy)(u+iv))=Re((xu)+i(yv))=yv=Im(z)Im(w)
  2. Take z=w=i. Then Re(z)=Re(w)=0 but Re(zw)=Re(1)=1. Also Im(z)=Im(w)=1 but Im(zw)=Im(1)=0.

    1. Calcuate z+w=(x+iy)+(u+iv)=(x+u)+i(y+v)=(x+u)i(y+v)=z+w

    2. Calcuate zw=(xuyv)+i(xv+yu)=(xuyv)i(xv+yu)=(xiy)(uiv)=zw

    3. Calcuate 1/z=xx2+y2iyx2+y2=xx2+y2+iyx2+y2=xx2+(y)2i(y)x2+(y)2=1/z

    4. z+z=(x+iy)+(xiy)=2x=2Re(z)
    5. zz=(x+iy)(xiy)=2iy=2iIm(z)
    1. The case n=1 is immediate. We assume (cosθ+isinθ)n=cos(nθ)+isin(nθ) and calculate (cosθ+isinθ)n+1=(cosθ+isinθ)n(cosθ+isinθ)=(cos(nθ)+isin(nθ))(cosθ+isinθ)=(cos(nθ+θ)+isin(nθ+θ))=cos((n+1)θ)+isin((n+1)θ) because when we multiply complex numbers we add arguments and multiply moduli.

    2. We have cos(3θ)+isin(3θ)=(cosθ+isinθ)3=(cosθ)3+3i(cosθ)2(sinθ)3(cosθ)(sinθ)2i(sinθ)3 so equating real and imaginary parts tell us cos(3θ)=(cosθ)33(cosθ)(sinθ)2 and sin(3θ)=3(cosθ)2(sinθ)(sinθ)3 hold.

      From (cosθ+isinθ)4=(cosθ)4+4i(cosθ)3(sinθ)6(cosθ)2(sinθ)24i(cosθ)(sinθ)3+(sinθ)4 we get also that cos(4θ)=(cosθ)46(cosθ)2(sinθ)2(sinθ)4 and sin(4θ)=4(cosθ)3(sinθ)4(cosθ)(sinθ)3 hold.

  3. Write z=r(cosθ+isinθ). If zn=w0 then |z|n=|w0| and nθ is an argument of w0. Thus nθ=Arg(w)+2kπ for some kZ. This means θ=Arg(w)+2kπn for some kZ. Taking k{0,1,,n1} gives all possible distinct z as after that we repeat.

  4. Take z1=z2=1+i3. We have Arg(z1)=Arg(z2)=2π3. But (1+i3)2=223i has principal argument 2π3.

Distance and Limits

  1. Fix ϵ>0. There is NN such that nN guarantees |zna|<ϵ. If nN then ||zn||a|||zna|<ϵ by the reverse triangle inequality.
    1. |zn|=|1+i|n=(2)n does not converge, so zn does not converge.
    2. Calculate |zn|=|(1+i)|nn=(2)nn and note that if n is large enough then (2)n/2>n so |zn|(2)n/2 which does not converge, so zn does not converge.

    3. |zn|=1|(1+i)n|=1(2)n converges to 0 so zn converges to 0.
  2. There is KN such that |z|/n<1 for all nK. Now for nK we have |znn!|=|z|1|z|2|z|K1|z|K|z|K+1|z|n|z|K!1K+11n which converges to zero as n.

Paths and Domains

  1. Fix 1c2. We calculate limtcγ(t)=limtc4t2+2it=4limtct2+2ilimtct=4c2+2ic=γ(c) using the limit laws, which proves γ is continuous at c. Since 1c2 was arbitrary γ is continuous on [1,2].

    1. Could be the image of a path.
    2. Could be the image of a path.
    3. Could not be the image of a path as it is not contiguous.
    4. Could be the image of a path.
  2. Take γ(t)=(1t)2+t(i1)=2+t(i3) which is continuous and satisfies has γ(0)=2 and γ(1)=i1.

    1. This is a domain as it is open and path connected.

    2. This is a domain as it is open and path connected.

    3. This is not a domain as it is not path connected.

    4. This is a domain as it is open and path connected.

  3. Fix zEF. Since E is open we can find r>0 such that B(z,r)E. Since F is open we can find s>0 such that B(z,s)F. Put t=min{r,s}. Then B(z,t)EF. Since zEF was arbitrary, the intersection is open.
  4. The red and blue sets above are both domains. However, their intersection is not path connected. The intersection of two domains need not be a domain.