# Thread: Write 1-i exponential form

1. ## Write 1-i exponential form

I have to write 1-i in exponential form. Please check my answer. If it is okay, yeah! If it is incorrect, could you help me?

r=sq.rt.(1^2 - (i)^2)

r=sq.rt.(1 + 1)

r=sq.rt.(2)

arctan 1/1 = pi/4 (45 degrees) cos pi/4 = .707, sin pi/4 = .707

.707/.707 =1

Is this correct?

2. Hello,

The problem is that it's arctan 1/(-1), because it's 1+(-1)i
so this gives the angle -pi/4

3. You have some mistaken ideas.
$\begin{gathered}
r = \left| {a + bi} \right| = \sqrt {a^2 + b^2 } \hfill \\
r = \left| {1 - i} \right| = \left| {1 + ( - 1)i} \right| = \sqrt {1^2 + ( - 1)^2 } = \sqrt 2 \hfill \\
\end{gathered}$
.

$\arg (1 - i) = \arctan \left( {\frac{{ - 1}}{1}} \right) = - \frac{\pi }{4}$.

So $1 - i = \sqrt 2 e^{\displaystyle\frac{{ - \pi }}{4}i}$

4. Originally Posted by Joanie
r=sq.rt.(1^2 - (i)^2)

Is this correct?
If $z = x+iy$, then $r = \sqrt{x^2 + y^2}$, so the above step is conceptually wrong...You should have written $r = \sqrt{1^2 + (-1)^2}$

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### what is the exponential form of 1-i

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