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jashcacheBest ResponseYou've already chosen the best response.2
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... \[r = \left Z \right = \sqrt{(\sqrt{3}1)^2 + i(\sqrt{3}+1)^2}\] \[\sqrt{(32\sqrt{3}+1)+(3+2\sqrt{3}+1)}\] \[= \sqrt{8}\] 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: \[Z = 2\sqrt{2}(\cos (75) + i \sin(75))\]
 2 years ago

Noureldin2022Best ResponseYou've already chosen the best response.0
\[(\sqrt{3}+i)(1+i)\] 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 z2sqrt(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)
 one year ago
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