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2 years ago

Find all complex solutions of the equation z⁴-1-i=0.

Mathematics

1 answer:

bulgar [2K]2 years ago

4 0

Answer:

Step-by-step explanation:

hello :

z⁴=1+i given : z² = t t is the complex number so : z⁴= (z²)² =t²

solve for t this equation : t² = 1+i

if : t = x+iy t² =(x+iy)² = x²+2xiy+(iy)² =x²-y²+2xyi ...... ( i² = -1)

t² = 1+i means : x²-y²+2xyi = 1+i

you have this system : x²-y² = 1.....(*)

2xy = 1.....(**)

slve for x and y you ca add this equation : ....(***)

( use : t² = 1+i

/t²/ = /1+i/ so :/t/² = /1+i/ .... (√(x²+y²))²=√(1²+1²) =√2 means x²+y² = √2)

the system is : x²-y² = 1.....(*)

2xy = 1.....(**)

x²+y² = √2 ....(***)

add (*) and (***) : 2x²= 1+√2

x² = (√2+1)/2

substrac (*) and (***) you have : y² = (√2-1)/2

use (**) the proudect xy is positif (same sign) so :

x=√( (√2+1)/2) and y = √( (√2-1)/2) so : t1 =√( (√2+1)/2)+i√( (√2-1)/2)

x= -√( (√2+1)/2) and y = -√( (√2-1)/2) so :t2= -t1

same method for equation : z² =t1 ..(2 solution ) and z² = t2..(2 solution )

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mariarad [96]

Answer:

Total number of ways will be 209

Step-by-step explanation:

There are 6 boys and 4 girls in a group and 4 children are to be selected.

We have to find the number of ways that 4 children can be selected if at least one boy must be in the group of 4.

So the groups can be arranged as

(1 Boy + 3 girls), (2 Boy + 2 girls), (3 Boys + 1 girl), (4 boys)

Now we will find the combinations in which these arrangements can be done.

1 Boy and 3 girls = $^{6}C_{1}\times^{4}C_{3}=6\times4$ =24

2 Boy and 2 girls= $^{6}C_{2}\times^{4}C_{2}=\frac{6!}{4!\times2!}\times\frac{4!}{2!\times2!}=15\times6=90$

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earnstyle [38]

Let's see

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Proved

Fact:-

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aleksandrvk [35]

Answer:

Step-by-step explanation:

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Where

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r represents the common ratio.

n represents the number of terms.

From the information given,

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fgiga [73]

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3 years ago

Find all complex solutions of the equation z⁴-1-i=0.​

Find all complex solutions of the equation z⁴-1-i=0.