a. ∣3−4i∣=9+16=5,arg(3−4i)=arctan(−4/3)≈−530|3-4i|=\sqrt{9+16}=5,arg(3-4i)=\arctan (-4/3)\approx-53^0∣3−4i∣=9+16=5,arg(3−4i)=arctan(−4/3)≈−530
b.∣−2+i∣=4+1=5,arg(−2+i)=arctan(−1/2)+1800≈153.40|-2+i|=\sqrt{4+1}=\sqrt5,arg(-2+i)=\arctan(-1/2)+180^0\approx153.4^0∣−2+i∣=4+1=5,arg(−2+i)=arctan(−1/2)+1800≈153.40
c.∣11+2i∣=∣1−i21+2∣=∣1/3−i2/3∣=1/9+2/9=3/3|\frac{1}{1+\sqrt2i}|=|\frac{1-i\sqrt2}{1+2}|=|1/3-i\sqrt2/3|=\sqrt{1/9+2/9}=\sqrt3/3∣1+2i1∣=∣1+21−i2∣=∣1/3−i2/3∣=1/9+2/9=3/3
arg11+i2=arctan(−2)≈54.70arg\frac{1}{1+i\sqrt2}=\arctan(-\sqrt2)\approx54.7^0arg1+i21=arctan(−2)≈54.70
d.∣7−i−3−3i∣=∣−(7−i)(1−i)3⋅2∣=∣−1+i4/3∣=1+16/9=5/3|\frac{7-i}{-3-3i}|=|-\frac{(7-i)(1-i)}{3\sdot2}|=|-1+i4/3|=\sqrt{1+16/9}=5/3∣−3−3i7−i∣=∣−3⋅2(7−i)(1−i)∣=∣−1+i4/3∣=1+16/9=5/3
arg(−1+i4/3)=arctan(−4/3)+1800≈1270arg(-1+i4/3)=\arctan(-4/3)+180^0\approx127^0arg(−1+i4/3)=arctan(−4/3)+1800≈1270
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments