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−49=\sqrt[]{-49}=−49= (a)
5−−100=5-\sqrt[]{-100}=5−−100= (a)
5i5+i3=5i^5+i^3=5i5+i3=
4i
-4i
-4
(1+i)+(2−i)=\left(1+i\right)+\left(2-i\right)=(1+i)+(2−i)= (a)
(1+i)(2−i)=\left(1+i\right)\left(2-i\right)=(1+i)(2−i)=
2+i2+i2+i
3+i3+i3+i
1+i1+i1+i
(3+i)-(1-i)= (a)
Solve 2−i1+i\frac{2-i}{1+i}1+i2−i and write answer in a+bi
12−32i\frac{1}{2}-\frac{3}{2}i21−23i
1−i2\frac{1-i}{2}21−i
−12i+12-\frac{1}{2}i+\frac{1}{2}−21i+21
Find argument for p=3-i
18.43°18.43\degree18.43°
161.57°161.57\degree161.57°
198.43°198.43\degree198.43°
Find modulus for p=3+i
10\sqrt[]{10}10
9\sqrt[]{9}9
7\sqrt[]{7}7
(2∠45°)×(5cos 333.43°+isin333.43°)\left(\sqrt[]{2}\angle45\degree\right)\times\left(\sqrt[]{5}\cos\ 333.43\degree+i\sin333.43\degree\right)(2∠45°)×(5cos 333.43°+isin333.43°)
10∠378.43°10\angle378.43\degree10∠378.43°
10∠288.43°\sqrt[]{10}\angle288.43\degree10∠288.43°
10∠378.43°\sqrt[]{10}\angle378.43\degree10∠378.43°
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