Find the arguments of $z_{1} = 5 + 5 i, z_{2} = - 4 + 4 i, z_{3} = - 3 - 3 i$ and $z_{4} = 2 - 2 i,$ where $i = \sqrt{- 1}$

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Solution

Since $z_{1}, z_{2}, z_{3}$ and $z_{4}$ lies in $I, I I, I I I$ and $I V$ quadrants respectively.The arguments are given by

1] arg $(z_{1}) = {tan}^{- 1} ∣ ∣ ∣ \frac{5}{5} ∣ ∣ ∣ = {tan}^{- 1} 1 = \frac{π}{4}$

2] arg $(z_{2}) = π - {tan}^{- 1} ∣ ∣ ∣ \frac{4}{- 4} ∣ ∣ ∣$

$= π - {tan}^{- 1} 1 = π - \frac{π}{4} = \frac{3 π}{4}$

3] arg $(z_{3}) = π + {tan}^{- 1} ∣ ∣ ∣ \frac{- 3}{- 3} ∣ ∣ ∣$

$= π + {tan}^{- 1} 1 = π + \frac{π}{4} = \frac{5 π}{4}$

and
4] arg $(z_{4}) = 2 π - {tan}^{- 1} ∣ ∣ ∣ \frac{- 2}{2} ∣ ∣ ∣$

$= 2 π - {tan}^{- 1} 1 = 2 π - \frac{π}{4} = \frac{7 π}{4}$

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Find the arguments of z1=5+5i,z2=−4+4i,z3=−3−3i and z4=2−2i, where i=√−1

Find the arguments of $z_{1} = 5 + 5 i, z_{2} = - 4 + 4 i, z_{3} = - 3 - 3 i$ and $z_{4} = 2 - 2 i,$ where $i = \sqrt{- 1}$