№1 вычислить1) (3+4i)(6-5i) 2) 5+i/-4+3i №2 запишите комплексное число в стандартном...

$(3+4i)(6-5i)=18+20+(24-15)i=38+9i$

$\dfrac {5+i}{-4+3i}=\dfrac{(5+i)(-4-3i)}{16+9}=\dfrac{-17-19i}{25}=-\dfrac{17}{25}-\dfrac{19}{25}i$

$2-2\sqrt3i=4(\frac12-\frac{\sqrt3}2i)=4(\cos(-\frac\pi3)+\sin(-\frac\pi3))$

$3-4i=5(\frac35-\frac45i)=5(\cos(\arctan(-\frac43))+\sin(\arctan(-\frac43))i)$

$4x^2-8x+5=0\\ 4(x^2-2x+1)+1=0\\ (x-1)^2=-\frac14\\ x=1\pm\frac i2$

$(\frac12-\frac{\sqrt3}2i)^5=\cos(-\frac53\pi)+i\sin(-\frac53\pi)=\cos\frac\pi3+i\sin\frac\pi3=\frac12+\frac{\sqrt3}2i$

$z^2=5-12i=9-12i-4=(3-2i)^2\\z=\pm(3-2i)$