You are watching: (1-i)^3
displaystyle-11+2i Explanation:In 2 stages: displaystyleleft( extStage 1: o ext left(1-2i ight)^2 ight)displaystyleleft(left(left(1-2i ight) ight)left(1-2i ight) ight) ...
displaystyle-32i Explanation:First compose this facility number in polar type and then use De Moivre : displaystyleleft(1-i ight)^10=left(sqrt2angle-fracpi4 ight)^10=left
Hint: 1-i=sqrt 2e^-ipi/4. To find the over equality you deserve to think about it geometrically and also use the recognized techniques: Or algebraically through proving that given any a,bin mathbb R, the ...
displaystyle-64 Explanation: displaystylez=1-iwill it is in in 4th quadrant of argand diagram. Important to keep in mind for as soon as we find the argument. displaystyler=sqrt1^2+left(-1 ight)^2=sqrt2 ...
create the facility number displaystyleleft(-1-i ight)^6 on the displaystylere^θ=rleft(cosθ+isinθ ight) kind (polar form) and also the displaystylea+ib ...
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https://socratic.org/questions/whrite-the-complex-number-1-i-6-on-the-re-r-cos-i-sin-form-polarform-and-the-a-i
1s2s2p Apr 10, 2018 we knowdisplaystylez=-1-i=re^i hetadisplaystyler=sqrtleft(-1 ight)^2+left(-1 ight)^2=sqrt2displaystyle heta= an^-1left(1 ight)=fracpi4 ...
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left< eginarray l l 2 & 3 \ 5 & 4 endarray ight> left< eginarray together l together 2 & 0 & 3 \ -1 & 1 & 5 endarray ight>
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