Question

Rog sa ma ajutati cu rezolvarea exercitiilor

Answer 1

Răspuns:

a)(1-2i)²-(2/3+√3i/2)=

(1-4i+4i²)-(2/3+√3/2)=

1-4i-4-2/3-√3i/2=

-3-2/3-4i-√3i/2=

-11/3-(8i+√3i)/2

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b)(3-i)/(2i-7)= amplifici fractia  cu   -2i-7=

(3-i)(-2i-7)/(2i-7)(-2i-7)=

(6i+2i²-21+7i)/(-7+2i)(-7-2i)=

(-2-21+13i)(49-4i²)=

(-23+13i)(49+4)=

(-23+13i)/53

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3.z=x+yi

1-z-z*z¬=i

z¬=x-yi

Inlocuiesti

1-x-yi-(1+iy)(1-iy)=i

1-x-yi-(1-y²i²)=i

1-x-yi-1+y²=i=.>

1-x+y²-yi=i=>

{1-x+y²=0

{y²=1=. > y=±1

Caz1   y= -1

1-x+(-1)²=0

1-x+1=0

2-x=0

x=2

Caz2  y=1

1-x+1²=0

1-x+1=0

x=2=>

z1=2+i

z2=2-i

Explicație pas cu pas: