## eli123 Group Title helppppp 2 years ago 2 years ago

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: $Z = 2\sqrt{2}(\cos (75) + i \sin(75))$
$(\sqrt{3}+i)(1+i)$ 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)