Question

helppppp

Answer 1

First some points to keep in mind:
\[r = \left| Z \right|= \sqrt{x^2 + y^2}\]
\[x=r \cos \theta\]
\[y=r \sin \theta\]
and \[Z = r \cos \theta + r \sin \theta\]


\[(\sqrt{3}+i)(1+i) \]
First we want to get the expression into a form that is more familiar.
FIOL...
\[= \sqrt{3}+i \sqrt{3}+i +i^{2}\]
\[= \sqrt{3} + i \sqrt{3} +i -1 \]
Collect like terms...
\[=\sqrt{3} -1 + i \sqrt{3} +i\]
\[=\sqrt{3} -1 + i (\sqrt{3} +1)\]
So that is a bit more recognizable...
\[(\sqrt{3} +1) + (i(\sqrt{3}-1))\]
We next find the modulus r. So, we take the square root of both terms squared...
\[r = \left| Z \right| = \sqrt{(\sqrt{3}-1)^2 + i(\sqrt{3}+1)^2}\]
\[\sqrt{(3-2\sqrt{3}+1)+(3+2\sqrt{3}+1)}\]
\[= \sqrt{8}\]
to get: \[r=2 \sqrt{2}\]

Next we need to find the angle of our complext number. Remember...
if we a graph of our complex numbers, \[x = (\sqrt{3}-1)\] and \[i = (\sqrt{3} +1)\], we would see that both numbers are positive ( \[x \approx 2.7\] and \[i \approx .7\] ) so they are in the first quadrant.
Since we also have both x and y coordinates we can find the angle by finding
\[\tan^{-1}( (\sqrt{3}+1)/(\sqrt{3}-1) ) = 5\pi/12 = 75^{0}\]

so combining that angle with our modulus and our original Polar coordinate Z, we get:

\[Z = 2\sqrt{2}(\cos (75) + i \sin(75))\]

Answer 2

\[(\sqrt{3}+i)(1+i)\]
first
z1=sqrt(3)+i
|z1|=sqrt(3+1)=2
z1=2(sqrt(3)/2+0.5i)
cos(theta)=sqrt(3)/2 .......... sin(theta)=0.5 -----> theta=30
z1=2*exp(30i)
second
z2=1+i
|z2|sqrt(1+1)=sqrt(2)
z2=sqrt(2)(1/sqrt(2)+1/sqrt(2)i)
cos(phi)=1/sqrt(2) ------ sin(phi)=1/sqrt(2) ------> phi=45
z2=sqrt(2)*exp(45i)
z=z1*z2
=\[2*\sqrt{2}*e ^{30i}*e ^{45i}\]
=\[2\sqrt{2}*e ^{75i}\]
=2sqrt(2)*(cos(75)+sin(75)*i)