Question

Please help it’s due tomorrow

Answer 1

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