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Príklady z Matematiky 3

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3 Tretí týµzdeµn<br />

1. V úlohách 1 - 5 zistite, aká mnoµzina je urµcená daným vz , tahom. Jej obraz<br />

naµcrtnite v komplexnej rovine.<br />

2. jz z 0 j = r; r > 0; z 0 je pevný bod. [Kruµznica so stredom z 0 a polomerom r]<br />

h<br />

i<br />

3. jz + ij + jz ij < 4: Vnútro elipsy x2<br />

3 + y2<br />

4 = 1<br />

4. jz + 2j > 1: [Vonkajšok kruµznice so stredom S = ( 2; 0) a polomerom r = 1]<br />

5. jz 2j < jzj : [Polrovina Re z > 1:]<br />

6. Im<br />

1<br />

z = 2: z 6= 0; kruµznica so stredom S = 0;<br />

<br />

1<br />

4 a polomerom r =<br />

1<br />

4<br />

7. Zistite, µci sú nasledujúce mnoµziny oblasti. (Naµcrtnite ich v komplexnej<br />

rovine):<br />

(a) jzj < 4; [áno]<br />

(b) 1 jz<br />

1j 3; [nie]<br />

(c) 3 arg z 3 4 ; [nie]<br />

(d) 0 < jz<br />

(e) Re z < 2: [áno]<br />

2j < 3; [áno]<br />

8. Nájdite limity postupnosti fz n g 1 n=1 ; ak<br />

<br />

(a) z n = 1 + 1 n<br />

3n +<br />

n+1<br />

3n 1 i; p 3 e +<br />

1<br />

3 i<br />

(b) z n = 2n sin 1 n + 4n+1<br />

5n 1 i; <br />

2 +<br />

4<br />

5 i<br />

(c) z n = n tg 1<br />

2n + 1 + 4 n<br />

n<br />

i;<br />

1<br />

2 + ie4<br />

9. Zistite, µci rady P 1<br />

n=1 z n konvergujú, alebo divergujú<br />

(a) z n =<br />

sin n+i cos n<br />

n 3<br />

; [absolútne konverguje]<br />

(b) z n = 1<br />

n(n+1) + tg <br />

2<br />

i; [absolútne konverguje]<br />

q<br />

n+1 h<br />

n+1<br />

(c) z n =<br />

n + n 3<br />

i; diverguje, návod rad P 1<br />

n n=1<br />

q<br />

n+1<br />

n<br />

10. Vyjadrite reálnu a imaginárnu µcas t , funkcie:<br />

h<br />

i<br />

(a) f (z) = e z2 ; Re f (z) = e x2 y 2 cos 2xy; Im f (z) = e x2 y 2 sin 2xy<br />

i<br />

nespĺµna nutnú podmienku konvergencie<br />

(b) f (z) = z 2 sin z; Re f (z) = x 2 y 2 sin x cosh y 2xy cos x sinh y; Im f (z) = 2xy sin x cosh y + x<br />

(c) f (z) = tg z; Re f (z) = x 2 y 2 sin x cosh y 2xy cos x sinh y; Im f (z) = 2xy sin x cosh y + x 2<br />

<br />

(d) f (z) = z 2 z + 1; Re f (z) = x<br />

2<br />

y 2 x + 1; Im f (z) = 2xy y <br />

(e) f (z) = 1 z ; hRe f (z) = x<br />

x 2 +y 2 ; Im f (z) =<br />

i<br />

y<br />

x 2 +y 2<br />

3

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