Question

what changes we make in different quadrants to find argument..give the difference
1)Convert the given complex number in polar form: 1 – i
2)Convert the given complex number in polar form: – 1 + i

Answer 1

@ganeshie8 help me plzz sir i have a test tomorrow

Answer 2

@mathmale

Answer 3

@surjithayer

Answer 4

@hartnn

Answer 5

@Kainui

Answer 6

\[1-\iota=r \left( \cos \theta+\iota \sin \theta \right)=r \cos \theta +\iota r \sin \theta\]
\[r \cos \theta=1,r \sin \theta=-1\]
square and add
\[r^2\left( \cos ^2\theta +\sin ^2 \theta \right)=1^2+\left( -1 \right)^2=2,r=\sqrt{2}\]
\[\sqrt{2}\sin \theta=-1,\sin \theta=-\frac{ 1 }{ \sqrt{2} }=-\sin \frac{ \pi }{ 4 }=\sin \left( \frac{ - \pi }{ 4 } \right)\]
\[=\sin \left( 2 \pi -\frac{ \pi }{ 4 } \right)=\sin \frac{ 7 \pi }{ 4 },\theta=\frac{ 7 \pi }{ 4 }\]
\[1-\iota =\sqrt{2}\cos \frac{ 7 \pi }{ 4 }+\iota \sqrt{2}\sin \frac{ 7 \pi }{ 4 }\]