Question

If |1 + i| x = 2 then write the value of x.

Answer 1

Appropriate Question :-

$\sf \: If \: { |1 + i| }^{x} = 2, \: then \: write \: the \: value \: of \: x$

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

Given that,

$\rm :\longmapsto\: { |1 + i| }^{x} = 2$

We know,

$\red{\boxed{\sf{ If \: z = x + iy, \: then \: |z| = \sqrt{ {x}^{2} + {y}^{2} }}}}$

So, using this identity, we get

$\rm :\longmapsto\: {( \sqrt{ {1}^{2} + {1}^{2} } )}^{x} = 2$

$\rm :\longmapsto\: {( \sqrt{1 + 1} )}^{x} = 2$

$\rm :\longmapsto\: {( \sqrt{2} )}^{x} = 2$

can be further rewritten as

$\rm :\longmapsto\: {( \sqrt{2} )}^{x} = {( \sqrt{2}) }^{2}$

$\bf\implies \:x = 2$

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Additional Information

Argument of complex number

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$

Answer 2

Answer:

Appropriate Question :-

$\sf \: If \: { |1 + i| }^{x} = 2, \: then \: write \: the \: value \: of \: x$

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

Given that,

$\rm :\longmapsto\: { |1 + i| }^{x} = 2$

We know,

$\purple{\boxed{\sf{ If \: z = x + iy, \: then \: |z| = \sqrt{ {x}^{2} + {y}^{2} }}}}$

So, using this identity, we get

$\rm :\longmapsto\: {( \sqrt{ {1}^{2} + {1}^{2} } )}^{x} = 2$

$\rm :\longmapsto\: {( \sqrt{1 + 1} )}^{x} = 2$

$\rm :\longmapsto\: {( \sqrt{2} )}^{x} = 2$

can be further rewritten as

$\rm :\longmapsto\: {( \sqrt{2} )}^{x} = {( \sqrt{2}) }^{2}$

$\bf\implies \:x = 2$

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