Title See discussions, stats, and author profiles for this publication at https www researchgate netpublication280722238 Complex Analysis Problems with solutions Book August 2016 CITATIONS 0 READS.
Basic algebraic and geometric properties 7
Proof Letz=x+yiwithx=Re(z)andy=Im(z) Then
Re(iz) =Re(−y+xi) =−y=−Im(z) and
Im(iz) =Im(−y+xi) =x=Re(z).
4 Verify the associative law for multiplication of complex numbers That is, show that
Proof Letz k =x k +iy k fork=1,2,3 Then
6 Let f be the map sending each complex number z=x+yi−→ x y
Proof Letz k =x k +y k ifork=1,2 Then z 1 z 2 = (x 1 +y 1 i)(x 2 +y 2 i) = (x 1 x 2 −y 1 y 2 ) +i(x 2 y 1 +x 1 y 2 ) and hence f(z 1 z 2 ) x 1 x 2 −y 1 y 2 x 2 y 1 +x 1 y 2
Sincei k = (−1) m fork=2meven andi k = (−1) m ifork=2m+1 odd,
8 Graph the following regions in the complex plane:
9 Find all complex solutions of the following equations:
(c) z= 9 z. Solution (a) Letz=z+iy Thus z = z x+iy = x+iy x−iy = x+iy
Hence,z=zif and only if Imz=0.
(b) Letz=z+iy Thus z+z = 0 x+iy+z+iy = 0 x−iy+x+iy = 0
Hence,z+zif and only if Rez=0.
(c) In this part we have z= 9 z ⇐⇒ zz=9 ⇐⇒ |z| 2 =9 ⇐⇒ |z|=3.
Hence,z= 9 z if and only if|z|=3
10 Suppose thatz 1 andz 2 are complex numbers, withz 1 z 2 real and non-zero Show that there exists a real numberrsuch thatz 1 =rz 2
Proof Letz 1 =x 1 +iy 1 andz 2 =x 2 +iy 2 withx 1 ,x 2 ,y 1 ,y 2 ∈R Thus z 1 z 2 =x 1 x 2 −y 1 y 2 + (x 1 y 2 +y 1 x 2 )i Sincez 1 z 2 is real and non-zero,z 1 6=0,z 2 6=0, and x 1 x 2 −y 1 y 2 6=0 and x 1 y 2 +y 1 x 2 =0.
Thus, sincez 2 6=0, then z 1 z 2 = x 1 +iy 1 x 2 −iy 2 ãx 2 +iy 2 x 2 +iy 2
By settingr=x 1 x 2 −y 1 y 2 x 2 2 +y 2 2 , we have the result
SinceQ⊂RandRis a field, we have the following:
Note thatb6=0 Thus we have
Modulus 13
Hint: Reduce this inequality to(|x| − |y|) 2 ≥0.
0≤(|Rez|+|Imz|) 2 =|Rez| 2 −2|Rez||Imz|+|Imz| 2
2|Rez||Imz| ≤ |Rez| 2 +|Imz| 2 , and then
|Rez| 2 +2|Rez||Imz|+|Imz| 2 ≤2(|Rez| 2 +|Imz| 2 ).
(|Rez|+|Imz|) 2 ≤2(|Rez| 2 +|Imz| 2 ) =2|z| 2 , and therefore,
3 Sketch the curves in the complex plane given by
(a) {Im(z) =−1}={y=−1}is the horizontal line passing through the point−i. (b) Since
⇔x+y=0, the curve is the linex+y=0.
= 4 3 the curve is the circle with centre at−2/3 and radius 4/3.
|z 2 +z+1| ≤ |z 2 |+|z|+|1|=|z| 2 +|z|+1=R 2 +R+1 by triangle inequality Hence z 4 +iz z 2 +z+1
Proof Since Log(z) =ln|z|+iArg(z)for−π