<h3>Answer:</h3><h3>40sqrt(2)*( cos(19pi/28) + i*sin(19pi/28) )</h3><h3>this can be written as </h3><h3>40sqrt(2)*cis(19pi/28)</h3>
The "cis" is shorthand for "cos + i*sin"
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Work Shown:
z1 = 5+5i is in the form a+bi with a = 5, b = 5
Use these values to find r and theta
r = sqrt(a^2+b^2)
r = sqrt(5^2+5^2)
r = sqrt(50)
r = sqrt(25*2)
r = sqrt(25)*sqrt(2)
r = 5sqrt(2)
theta = arctan(b/a)
theta = arctan(5/5)
theta = arctan(1)
theta = pi/4 radians
This tells us
z1 = 5+5i
turns into
z1 = 5sqrt(2)*(cos(pi/4)+i*sin(pi/4))
which is the polar form representation.
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The notation "cos(x)+i*sin(x)" is cumbersome to write down. Common shorthand convention is to use "cis", the letters of which stand for the first letter of "cosine", then the "i", then the first letter of "sine" in that order.
In other words,
cos(x)+i*sin(x) = cis(x)
cos(pi/4)+i*sin(pi/4) = cis(pi/4)
and so on. It makes using the rule shown later much easier to work with.
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What we really have are these two complex numbers
z1 = 5sqrt(2)*cis(pi/4)
z2 = 8*cis(3pi/7)
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Now we'll use this rule
if z1 = r1*cis(theta1) and z2 = r2*cis(theta2)
then z1*z2 = r1*r2*cis(theta1+theta2)
we multiply the magnitudes r1 and r2; while adding the angles theta1 and theta2
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So,
z1 = 5sqrt(2)*cis(pi/4)
z2 = 8*cis(3pi/7)
z1*z2 = 5sqrt(2)*8*cis(pi/4+3pi/7)
z1*z2 = 40sqrt(2)*cis(7pi/28+12pi/28)
z1*z2 = 40sqrt(2)*cis(19pi/28)
z1*z2 = 40sqrt(2)*( cos(19pi/28) + i*sin(19pi/28) )
The final r value is 40*sqrt(2)
The final theta value is 19pi/28 radians
To convert this into a+bi form, you would compute the sine and cosine values, distribute then simplify. I find it easier to keep it in polar form.
