Solve for z
z=1-i
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z≔1-i
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z=\frac{\left(7+i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}
Multiply both numerator and denominator of \frac{7+i}{3+4i} by the complex conjugate of the denominator, 3-4i.
z=\frac{\left(7+i\right)\left(3-4i\right)}{3^{2}-4^{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{\left(7+i\right)\left(3-4i\right)}{25}
By definition, i^{2} is -1. Calculate the denominator.
z=\frac{7\times 3+7\times \left(-4i\right)+3i-4i^{2}}{25}
Multiply complex numbers 7+i and 3-4i like you multiply binomials.
z=\frac{7\times 3+7\times \left(-4i\right)+3i-4\left(-1\right)}{25}
By definition, i^{2} is -1.
z=\frac{21-28i+3i+4}{25}
Do the multiplications in 7\times 3+7\times \left(-4i\right)+3i-4\left(-1\right).
z=\frac{21+4+\left(-28+3\right)i}{25}
Combine the real and imaginary parts in 21-28i+3i+4.
z=\frac{25-25i}{25}
Do the additions in 21+4+\left(-28+3\right)i.
z=1-i
Divide 25-25i by 25 to get 1-i.
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y = 3x + 4
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Simultaneous equation
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Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
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Limits
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