Solve for z
z=\frac{1}{5}+\frac{2}{5}i=0.2+0.4i
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z≔\frac{1}{5}+\frac{2}{5}i
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z=\frac{i\left(2-i\right)}{\left(2+i\right)\left(2-i\right)}
Multiply both numerator and denominator of \frac{i}{2+i} by the complex conjugate of the denominator, 2-i.
z=\frac{i\left(2-i\right)}{2^{2}-i^{2}}
Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
z=\frac{i\left(2-i\right)}{5}
By definition, i^{2} is -1. Calculate the denominator.
z=\frac{2i-i^{2}}{5}
Multiply i times 2-i.
z=\frac{2i-\left(-1\right)}{5}
By definition, i^{2} is -1.
z=\frac{1+2i}{5}
Do the multiplications in 2i-\left(-1\right). Reorder the terms.
z=\frac{1}{5}+\frac{2}{5}i
Divide 1+2i by 5 to get \frac{1}{5}+\frac{2}{5}i.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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4 \sin \theta \cos \theta = 2 \sin \theta
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y = 3x + 4
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Matrix
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Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
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