Question

exponential form of Z= 1+i is =​

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

The exponential form of a complex number is given by

$\sf \: z = r {e}^{i\theta\: } where$

$\rm :\longmapsto\: {e}^{i\theta} = cos\theta \: + i \: sin\theta$

$\rm :\longmapsto\:r \: = \: length \: of \: complex \: number$

$\rm :\longmapsto\:\theta \: is \: the \: argument \: of \: a \: complex \: number$

Let's solve the problem now!!

Now,

• Given complex number is 1 + i.

$\rm :\longmapsto\:Let \: 1 + i = r {e}^{i\theta} - - - (1)$

$\rm :\longmapsto\: \: 1 + i = r (cos\theta \: + i \: sin\theta)$

$\rm :\longmapsto\:1 + i = r cos\theta \: + \: i \: r \: sin\theta$

• On comparing, we get

$\rm :\longmapsto\: r cos\theta \: = \: 1 - - - (2)$

and

$\rm :\longmapsto\: r sin\theta \: = \: 1 - - - (3)$

• On squaring equation (1) and (2), and add we get

$\rm :\longmapsto\: { r }^{2} {cos}^{2} \theta \: + \: { r }^{2} {sin}^{2} \theta = 1 + 1$

$\rm :\longmapsto\: { r }^{2} ( {sin}^{2} \theta + {cos}^{2} \theta) = 2$

$\rm :\longmapsto\: { r }^{2} = 2 \: \: \: \: \: \: \: \{ \bf\because \: {sin}^{2} \theta + {cos}^{2} \theta = 1 \}$

$\bf\implies \: r = \sqrt{2} - - (4) \: \: as \: r \nless 0$

On substituting the value from (4) to equation (3) and (2),

$\rm :\longmapsto\:cos\theta = \dfrac{1}{ \sqrt{2} } \: \: and \: \: sin\theta = \dfrac{1}{ \sqrt{2} }$

$\sf \: as \: cos\theta > 0 \: and \: sin\theta > 0 \: it \: means \: \theta \in \: {1}^{st} \: quadrant$

$\bf\implies \:\theta \: = \: \dfrac{\pi}{4} - - (5)$

Now,

• Substituting the values from (4) and (5) in equation (1),

$\bf\implies \:1 + i = \sqrt{2} \: {e} \: ^ {i\dfrac{\pi}{4} }$

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Additional Information :-

The exponential form of a complex number is extension of its polar form. We kmow that the polar form of a complex number z is:

$\rm :\longmapsto\:z \: = \: r cos\theta + i \: sin\theta \: = \: r \: cis \: \theta$

Euler's Formula tells us that:

$\rm :\longmapsto\: {e}^{i\theta} = cos\theta \: + \: i \: sin\theta$

Thus,

• we can write

$\rm :\longmapsto\:z \: = \: r \: {e}^{i\theta}$

Explore more :-

$\rm :\longmapsto\: 1). \: \: {e}^{ - i \: \theta} = cos\theta \: - \: i \: sin\theta$

$\rm :\longmapsto\:2). \: \: cos\theta \: = \dfrac{ {e}^{i\theta \:} + {e}^{ - i\theta} }{2}$

$\rm :\longmapsto\:3). \: \: sin\theta \: = \dfrac{ {e}^{i\theta} - {e}^{ - i\theta} }{2i}$