Given z1= 1+I/√3-i , z2=e^1+(π/2) and z3=4(cos30°-isin30°) Find: (a) Z1 bar +Z3 (b) (Z1 - Z2 bar) / |Z1|+Z3 (c) Z1^5 +Z3^4 (d) Z1^2 / Z2^3 (e) |Z2 + Z1^2 |
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1 Answer

First convert the complex numbers into a+ib form:

z₁=1+i/(√3-i)=1+i(√3+i)/(3-i²)=1+¼i√3-¼=¾+¼i√3.

z̄₁=¾-¼i√3; |z₁|=√((9/16)+(3/16))=½√3.

z₂=(e^i)+π/2 (assumed) = π/2+cos(1)+isin(1). The angle is radians.

cos(1)=0.5403, sin(1)=0.8415 approx. π/2=1.5708 approx.

z₂=2.1111+0.8415i approx; z̄₂=2.1111-0.8415i.

z₃=2√3-2i; z̄₃=2√3+2i.

(a) z̄₁+z₃=¾-¼i√3+2√3-2i=(¾+2√3)-(2+¼√3)i=4.2141-2.4330i.

 

(b) z₁-z̄₂=¾+¼i√3-(2.1111-0.8415i)=-1.3611-0.4085i.

|z₁|+z₃=½√3+2√3-2i=5(√3)/2-2i;

1/(|z₁|+z₃)=(5(√3)/2+2i)/(91/4)=(10√3+8i)/91.

(z₁-z̄₂)/(|z₁|+z₃)=-(1.3611+0.4085i)(10√3+8i)/91=

-(1/91)(23.5749+17.9635i-3.2677)=-(1/91)(20.3073+17.9635i)=

-0.2232-0.1974i.

 

(c) z₁⁵+z₃⁴.

z₃ can be expressed 4e^(πi/6), so z₃⁴=256e^(⅔πi)=

256(cos(⅔π)+isin(⅔π))=256(-½+i√3/2)=

-128+128i√3.

z₁ can be expressed (√3/2)(cos30°+isin30°)=(√3/2)e^(πi/6).

z₁⁵=9√3/32e^(⅚πi)=(9√3/32)(-cos30°+isin30°)=

-27/64+9i√3/64.

z₁⁵+z₃⁴=-27/64+9i√3/64-128+128i√3=

-8219/64+221.9461i approx.

 

(d) z₁²=9/16-3/16+3i√3/8=⅜+⅜i√3.

1/z₂=1/(π/2+cos(1)+isin(1))=

(π/2+cos(1)-isin(1))/((π/2+cos(1))²+sin²(1))=

(π/2+cos(1)-isin(1))/(π²/4+πcos(1)+1)=

(π/2+cos(1)-isin(1))/5.1648=

0.4087-0.1629i.

1/z₂³=0.4087³-3(0.4087)²(0.1629)i+3(0.4087)(0.1629)²i²-(0.1629)³i³

0.0683-0.0817i-0.0325+0.0043i=

0.0357-0.0773i approx.

z₁²/z₂³=(⅜+⅜i√3)(0.0357-0.0773i)=

0.0134-0.0290i+0.0232-0.0502=0.0366-0.0792i.

 

(e) z₂+z₁²=2.1111+0.8415i+⅜+⅜i√3=

2.4861+1.4910i.

|z₂+z₁²|=√(2.4861²+1.4910²)=2.8989.

by Top Rated User (721k points)

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