Answer: 10*(cos(pi) + i*sin(pi))

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Work Shown:

z1 = 1+i is in the form a+bi where a = 1 and b = 1
r = sqrt(a^2+b^2)
r = sqrt(1^2+1^2)
r = sqrt(2)
theta = arctan(b/a)
theta = arctan(1/1)
theta = pi/4

The polar form for z1 is
z1 = r*(cos(theta) + i*sin(theta))
z1 = sqrt(2)*(cos(pi/4)+i*sin(pi/4)

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z2 = -5+5i is in the form a+bi where a = -5 and b = 5
r = sqrt(a^2+b^2)
r = sqrt((-5)^2+5^2)
r = sqrt(50)
r = 5*sqrt(2)
theta = arctan(b/a)
theta = arctan(5/(-5))
theta = 3pi/4

The polar form for z2 is
z2 = r*(cos(theta) + i*sin(theta))
z2 = 5*sqrt(2)*(cos(3pi/4)+i*sin(3pi/4)
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Now use the rule
If
z = a*(cos(b) + i*sin(b)) and w = c*(cos(d)+i*sin(d))
then
z*w = a*c*(cos(b+d)+i*sin(b+d))

We have
a = sqrt(2)
b = pi/4
c = 5*sqrt(2)
d = 3pi/4

So...

z*w = a*c*(cos(b+d)+i*sin(b+d))
z1*z2 = sqrt(2)*5*sqrt(2)*(cos(pi/4+3pi/4)+i*sin(pi/4+3pi/4))
z1*z2 = 10*(cos(4pi/4)+i*sin(4pi/4))
z1*z2 = 10*(cos(pi)+i*sin(pi))
which is the answer in polar form

7 0

4 years ago

X2 – 6 3x + 10 ? g(4) = Be sure to simplify your answer. g(x) =

Triss [41]

5/11

((4)^2 - 6) / 3(4)+10
16 - 6 / 12 + 10
10/22 we can simplify this further
10 divided by 2
22 divided by 2

You get 5/11 which is the same as 10/22

8 0

2 years ago

PLEASE BRILLIANT i need help asap

Whitepunk [10]

Answer:

p = 6, q = - 1

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $(a^{m}) ^{n}$ = $a^{mn}$